= Solution
{c}
For $S=\sum_iX_i$, the ratio of the two sample <likelihood functions> is
$$
\frac{L(\mu_1)}{L(\mu_0)}
=e^{-n(\mu_1-\mu_0)}\left(\frac{\mu_1}{\mu_0}\right)^S.
$$
For $0<\mu_0<\mu_1$ it is strictly increasing in $S$. The <Neyman-Pearson lemma> therefore selects an upper tail:
$$
\boxed{C=\{\mathbf x:\textstyle\sum_i x_i\geq k\}.}
$$
Under the null, $S\sim\operatorname{Poisson}(n\mu_0)$, so choose $k$ with this upper-tail probability at most $\alpha$. If exact size $\alpha$ is required, randomize at the adjacent boundary value as needed. The rejection rule is the same for every $\mu_1>\mu_0$, giving a <uniformly most powerful test> for the one-sided alternative.
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