= Solution
A level-$\alpha$ test is <uniformly most powerful test>[uniformly most powerful] for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.
For every fixed $\lambda_1<\lambda_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $nc_\alpha$, independent of $\lambda_1$. The <Neyman-Pearson lemma> therefore makes the test from part (a) most powerful against every individual $\lambda_1<\lambda_0$. Consequently it is
$$
\boxed{\text{uniformly most powerful for }H_0:\lambda=\lambda_0
\text{ against }\widetilde H_1:\lambda<\lambda_0}.
$$
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