= Solution
A zero-cost stock holding $\theta$ has terminal payoff $\theta(S_1-1)$. If $a\le1$, shorting the share and holding a bond gives $1-S_1\ge0$ and a strictly positive payoff <almost surely>. If $a>1$, both signs of $S_1-1$ occur with positive <probability>, so no nonzero holding is an arbitrage. Hence \b[arbitrage freedom is equivalent to $a>1$].
The <equivalent martingale measures> are exactly the densities $Z(s)>0$ <almost everywhere> on $(0,a)$ satisfying
$$
\boxed{\frac1a\int_0^a Z(s)\,ds=1,\qquad\frac1a\int_0^a sZ(s)\,ds=1.}
$$
They exist for $a>1$: exponential tilts of the uniform law have means ranging continuously from zero to $a$, so one has mean one.
Under the usual utility assumptions $U'>0$, $U''<0$, and existence of an admissible optimum, $J(\theta)=E[U(w+\theta(S_1-1))]$ is <strictly concave> and $J'(0)=U'(w)(a/2-1)$. Its decreasing derivative implies that the optimizer is positive exactly when $a>2$, negative when $a<2$, and zero when $a=2$, provided the feasible interval contains zero in its interior. The hypothesis $C^2$ alone is insufficient: a linear increasing utility can give an unbounded objective rather than a finite optimizer. Concavity and existence are implicit in the standard utility formulation.
For $a=2$, $\theta=0$ makes marginal utility constant, so the pricing <measure> is the original uniform law and
$$
\boxed{p((S_1-1)^+)=\frac12\int_1^2(s-1)\,ds=\frac14.}
$$
For any equivalent pricing <measure>, the call <expectation> is positive and strictly less than $E_QS_1/2=1/2$, since $(s-1)^+<s/2$ for $0<s<2$. Both endpoints can be approached. Symmetric densities concentrated in $[1-\delta,1+\delta]$ give prices tending to zero; symmetric densities concentrated near zero and two give prices tending to one half. Mix either density with an arbitrarily small amount of the uniform law to keep it everywhere positive; symmetry keeps its mean one. The set of positive mean-one densities is convex and the call price is linear, so every intermediate value is attained. Thus \b[the exact price range is $(0,1/2)$], with neither endpoint attained.
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