Extras

For the first, use that $$\begin{align*} {\left\lvert { \sum_{k\geq 1} {1\over k^2} \sin^n(k) } \right\rvert} \leq \sum_{k\geq 1} {\left\lvert { {1\over k^2} \sin^n(k) } \right\rvert} \sum_{k\geq 1} {\left\lvert { {1\over k^2}} \right\rvert} < \infty ,\end{align*}$$ since ${\left\lvert {\sin(x)} \right\rvert} \leq 1$ and $x^n < x$ for ${\left\lvert {x} \right\rvert}\leq 1$. By the dominated convergence theorem, we can pass the limit inside. Using the same fact as above, $\lim_{n\to\infty}\sin^n(x) = 0$,

For the second, the claim is that it diverges (very slowly). Note that $\lim_{n\to\infty} e^{-k/n} = 1$ for any $k$. By Fatou, we have $$\begin{align*} \liminf_{n\to\infty} \sum_{k\geq 1} {e^{-k/n} \over k} \geq \sum_{k\geq 1} \liminf_{n\to\infty} {e^{-k/n} \over k} = \sum_{k\geq 1} {1 \over k} = \infty .\end{align*}$$

Apply Fubini-Tonelli to commute two sums: $$\begin{align*} \mu\qty{\bigcup_{1\leq k \leq M} E_k}\coloneqq &= \sum_{n\geq 1} \mu_n\qty{\bigcup_{1\leq k \leq M} E_k}\\ &= \sum_{n\geq 1} \sum_{1\leq k \leq M} \mu_n\qty{E_k}\\ &= \sum_{1\leq k \leq M}\sum_{n\geq 1} \mu_n\qty{E_k} \text{FT} \\ &\coloneqq\sum_{1\leq k \leq M} \mu(E_k) .\end{align*}$$