Q2: No. Counterexample: Suppose there's one outcome $x$ such that all lotteries are equally good, except for the lottery than puts probability 1 on $x$, which is worse than the others.

Q5 is true if (as you assumed), the space of lotteries is the space of distributions over a finite set. (For a general convex set, you can get long-line phenomena.)

First, without proof, I'll state the following generalization.

Theorem 1. Let $⪯$ be a relation on a convex space $L$ satisfying axioms A1, A2, A3, and the following additional continuity axiom. For all $A,B_1,B_2,C\in L$, the set

is open in $[0,1]$. Then, there exists a function $u$ from $L$ to the long line such that $u\left(A\right)\le u\left(B\right)$ iff $A⪯B$.

The proof is not too different, but simpler, if we also assume A4. In particular, we no longer need the extra continuity axiom, and we get a stronger conclusion. Nate sketched part of the proof of this already, but I want to be clearer about what is stated and skip fewer steps. In particular, I'm not sure how Nate's hypotheses rule out examples that require long-line-valued functions—maybe he's assuming that the domain of the preference relation is a finite-dimensional simplex like I am, but none of his arguments use this explicitly.

Theorem 2. Let $⪯$ be a relation on a finite-dimensional simplex $L=\Delta \Omega$ satisfying axioms A1-A4. Then, there is a quasiconcave function $u:L\to R$ such that $u\left(A\right)\le u\left(B\right)$ iff $A⪯B$.

First, I'll set up some definitions and a lemma. For any lotteries $A$, $B$, let $[A,B]$ denote the line segment 

We say that preferences are increasing along a line segment $[A,B]$ if whenever $p\le q$, we have

We will also use open and half-open interval notation in the corresponding way.

Lemma. Let $⪯$ be a preference relation on a finite-dimensional simplex $L=\Delta \Omega$ satisfying axioms A1-A4. Then, there are $⪯$-minimal and -maximal elements in $L$.

Proof. First, we show that there is a minimal element. Axiom A4 states that for any mixture $C=pA+(1-p)B$, either $C⪰A$ or $C⪰B$. By induction, it follows more generally that any convex combination C of finitely many elements ${\left(A_i\right)}_{i\in I}$ satisfies $C⪰A_i$ for some $i\in I$. But every element is a convex combination of the vertices of $L$, so some vertex of $L$ is $⪯$-minimal.

The proof that there is a maximal element is more complex. Consider the family of sets

This is a prefilter, so since $L$ is compact ($L$ here carries the Euclidean metric), it has a cluster point $B$. Either $B$ will be a maximal element, or we will find some other maximal element. In particular, take any $A\in L$. We are done if $A$ is a maximal element; otherwise, pick $A^{\prime }\succ A$. By the construction of $F$, for every $n\in N$, we can pick some $C_n⪰A^{\prime }$ within a distance of $\frac{1}{n}$ from B. Now, if we show that $B$ itself satisfies $B⪰A$, it will follow that $B$ is maximal.

The idea is to pass from our sequence ${\left(C_n\right)}_{n\in N}$, with limit $B$, to another sequence lying on a line segment with endpoint $B$. We can use axiom A4, which is a kind of convexity, to control the preference relation on convex combinations of our points $C_n$, so these are the points that we will construct along a line segment. Once we have this line segment, we can finish by using A3, which is a kind of continuity restricted to line segments, to control $B$ itself.

Let $S\subseteq L$ be the set of lotteries in the affine span of the set ${\left\{C_n\right\}}_{n\in N}$. Then, if we take some index set $I\subseteq N$ such that ${\left(C_n\right)}_{n\in I}$ is a maximal affinely independent tuple, it follows that ${\left\{C_n\right\}}_{n\in I}$ affinely generates $S$. Hence, the convex combination

i.e. the barycenter of the simplex with vertices at ${\left(C_n\right)}_{n\in I}$, is in the interior of the convex hull of ${\left\{C_n\right\}}_{n\in I}$ relative to $S$, so we can pick some $r>0$ such that the $r$-ball around $D$ relative to $S$ is contained in this simplex.

Now, we will see that every lottery $E$ in the set $(B,D]$ satisfies $E⪰A^{\prime }$. For any $\epsilon >0$, pick $k$ so that $C_k$ is in the $\epsilon$-ball around $B$. Since the tangent vector $v=B-C_k$ has length less than $\epsilon$, the lottery

is in the $r$-ball around $D$, and it is in $S$, so it is in the simplex with vertices ${\left(C_n\right)}_{n\in I}$. Then, $F⪰A^{\prime }$ by A4, and $C_k⪰A^{\prime }$ by hypothesis. So, applying A4 again,

Using A4 one more time, it follows that every lottery

satisfies $E⪰A^{\prime }$, and hence every lottery $E\in (B,D]$.

Now we can finish up. If $B\prec A$ then, using A3 and the fact that $D⪰A^{\prime }\succ A$, there would have to be some lottery in $[B,D]$ that is $⪯$-equivalent to A, but this would contradict what we just concluded. So, $B⪰A$, and so B is $⪯$-maximal. $\square$

Proof of Theorem 2. Let $C$ be a $⪯$-minimal and $D$ a $⪯$-maximal element of $L$. First, we will see that preferences are increasing on $[C,D]$, and then we will use this fact to construct a function $L\to R$ and show that it has the desired properties. Suppose preferences we not increasing; then, there would be $A,B\in [C,D]$ such that $A$ is closer to $C$ while $B$ is closer to $D$, and $A\succ B$. Then, $B$ would be a convex combination of $A$ and $D$, but $B\prec A⪯D$ by the maximality of $D$, contradicting A4.

Now we can construct our utility function $u:L\to R$ using A3; for each $\sim$-class $\left[A\right]$, we have $C⪯A⪯D$, so there is some^[1] $p\in [0,1]$ such that

Then, let $u\left(A^{\prime }\right)=p$ for all $A^{\prime }\in \left[A\right]$. Since preferences are increasing on $[C,D]$, it is immediate that if $u\left(A\right)\le u\left(B\right)$, then $A⪯B$. Conversely, if $A⪯B$, we have two cases. If $A\prec B$, then $B⋠A$, so $u\left(B\right)\nleqq u\left(A\right)$, and so $u\left(A\right)\le u\left(B\right)$. Finally, if $A\sim B$, then $u\left(A\right)=u\left(B\right)$ by construction.

Finally, since for all $A,B\in L$ we have $u\left(A\right)\le u\left(B\right)$ iff $A⪯B$, it follows immediately that $u$ is quasiconcave by A4. $\square$

1. ^{^}

   Nate mentions using choice in his answer, but here at least the use of choice is removable. Since $⪯$ is monotone on $[C,D]$, the intersection of the $\sim$-class $\left[A\right]$ with $[C,D]$ is a subinterval of $[C,D]$, so we can pick $p$ based on the midpoint of that interval

The answers to Q3, Q4 and Q6 are all no. I will give a sketchy argument here.

Consider the one dimensional case, where the lotteries are represented by real numbers in the interval $L=[0,1]$, and consider the function $u:L\to [0,1]$ given by $u\left(x\right)=\frac{1}{2}-(x-\frac{1}{3}{)}^3(x-\frac{2}{3})$. Let $⪰$ be the preference order given by $x⪰y$ if and only if $u\left(x\right)\ge u\left(y\right)$.

$u$ is continuous and quasi-concave, which means $⪰$ is going to satisfy A1, A2, A3, A4, and B2. Further, since $u$ is monotonically increasing up to the unique argmax, and then monotonically decreasing, $⪰$ is going to satisfy A5. 

$u$ is not concave, but we need to show there is not another concave function giving the same preference relation as $u$. The only way to keep the same preference relation is to compose $u$ with a strictly monotonic function $f$, so $v\left(x\right)=f\left(u\right(x)$).

If $f$ is smooth, we have a problem, since $v^{\prime }\left(\frac{1}{3}\right)=f^{\prime }\left(u\right(\frac{1}{3}\left)\right)u^{\prime }\left(\frac{1}{3}\right)=f^{\prime }\left(\frac{1}{2}\right)0=0$. However, since, $v^{\prime }$ must be on some $x>\frac{1}{3}$, but concavity would require $v^{\prime }$ to be decreasing.

In order to remove the inflection point at $x=\frac{1}{3}$, we need to flatten it out with some $f$ that has infinite slope at $\frac{1}{2}$. For example, we could take $f\left(z\right)=\sqrt[3]{3z-\frac{1}{2}}$. However, any f that removes the inflection point at $x=\frac{1}{3}$, will end up adding an inflection point at $x=\frac{2}{3}$, which will have a infinite negate slope. This newly created inflection point will cause a problem for similar reasons.

The answer to Q1 is no, using the same counter example here. However, the spirit of my original question lives on in Q4 (and Q6).

Claim: A1, A2, A3, A5, and B2 imply A4.

Proof: Assume we have a preference ordering that satisfies A1, A2, A3, A5, and B2, and consider lotteries $A,B\in L$, and $p\in [0,1]$, with $A⪰B$. Let $C=pA+(1-p)B$. It suffices to show $C⪰B$. Assume not, for the purpose of contradiction. Then (by axiom A1), $B\succ C$. Thus by axiom B2 there exists a $D\in L$ such that $B\succ D\succ C$. By axiom A3, we may assume $D=qB+(1-q)C$ for some $q\in [0,1]$. Observe that $C=rA+(1-r)D$ where $r=\frac{pq}{1-p+pq}\in [0,1]$. $r$ is positive, since otherwise $C=D\succ C$. Thus, we can apply A5 to get that since $D⪰rA+(1-r)D$, we have $D⪰A$. Thus $D⪰A⪰B\succ D$, a contradiction.

No on Q4? I think Alex's counterexample applies to Q4 as well.

(EDIT: Scott points out I'm wrong here, Alex's counterexample doesn't apply, and mine violates A5.)

In particular I think A4 and A5 don't imply anything about the rate of change as we move between lotteries, so we can have movements too sharp to be concave. We only have quasi-concavity.

My version of the counterexample: you have two outcomes $X$ and $\neg X$, we prefer anything with $P\left(X\right)\le \frac{1}{2}$ equally, and we otherwise prefer higher $P\left(X\right)$.

If you give me a corresponding $u\left(p\right)$, it must satisfy $u\left(0\right)=u\left(\frac{1}{2}\right)<u\left(1\right)$, but convexity demands that $u\left(\frac{1}{2}\right)\ge \frac{1}{2}u\left(0\right)+\frac{1}{2}u\left(1\right)$, which in this case means $u\left(0\right)\ge u\left(1\right)$, a contradiction.