= Solution
Let $K=\sup_T\mathbb E[e^{M_T/2}]<\infty$, with $T$ ranging over finite stopping times. To locate the <half-threshold for the exponential-martingale Hölder bound>, put $u=\sqrt{pq}>\sqrt q$. Since $u(u-1)$ is increasing for $u>1$,
$$
C(p,q)>\frac{q-\sqrt q}{q-1}
=\frac{\sqrt q}{\sqrt q+1}>\frac12.
$$
On the other hand, taking $q=1+\varepsilon$ and $p=1+\varepsilon^2$ gives $C(p,q)\to1/2$ as $\varepsilon\downarrow0$. Hence
$$
\boxed{\inf_{p,q>1}C(p,q)=\frac12.}
$$
Fix $0<a<1$. Choose $p,q>1$ with $aC(p,q)<1/2$, and put $\lambda=2aC(p,q)\in(0,1)$. Part (a) and the <Jensen inequality> give, for every finite stopping time,
$$
\begin{aligned}
\mathbb E[\mathcal E(aM)_T^p]
&\leq\bigl(\mathbb E[(e^{M_T/2})^\lambda]\bigr)^{(q-1)/q}\\
&\leq K^{\lambda(q-1)/q}.
\end{aligned}
$$
Thus the entire stopped family has a uniform $L^p$ bound for some $p>1$. By <uniform integrability from an Lp bound>, it is <uniformly integrable>.
To check that this <local martingale> is a true <martingale>, stop it by a <localizing sequence>. At each fixed time the stopped variables are <uniformly integrable> by the same bound; taking limits in their conditional <martingale> identities proves the unstopped identity. The uniform bound over all finite stopping times then makes it a <uniformly integrable martingale>, with terminal expectation one. Therefore
$$
\boxed{\mathcal E(aM)\text{ is a uniformly integrable martingale for every }0<a<1.}
$$
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