Assuming $E^n = F_n$.
To show $h \in L^p$ you need to show $(\int_X |h|^p )^{1/p} < \infty$:
$$ \left( \int_X |h|^p \right)^{1/p} = \left( \int_X |\sum \chi_{F_n}|^p \right)^{1/p} \leq \left( \int_X \left( \sum |\chi_{F_n}| \right)^p \right)^{1/p}$$
Then using the Lebesgue dominated convergence theorem you know that you can do this:
$$ \left( \int_X \left( \sum |\chi_{F_n}| \right)^p \right)^{1/p} = \left( \int_X \lim_{N \rightarrow \infty} \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p \right)^{1/p} = \left(\lim_{N \rightarrow \infty} \int_X \left(\sum_{n=1}^N |\chi_{F_n}| \right)^p \right)^{1/p}$$
i.e. you can swap limit and integral and then because you have a finite sum you can swap integral and sum. Let's put things together:
$$ \begin{align*}
\left( \int_X |h|^p \right)^{1/p} = \left( \int_X \Big |\sum_{n=1}^\infty \chi_{F_n} \Big |^p dm \right)^{1/p} \\
\stackrel{\Delta-ineq.}{\leq} \left( \int_X \left( \sum_{n=1}^\infty |\chi_{F_n}| \right)^p dm \right)^{1/p} \\
= \left( \int_X \left( \lim_{N \rightarrow \infty} \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\
\stackrel{cont. of ()^p}{=} \left( \int_X \lim_{N \rightarrow \infty} \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\
\stackrel{Lebesgue}{=} \left( \lim_{N \rightarrow \infty} \int_X \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\
\stackrel{\|.\| \Delta ineq.}{\leq} \lim_{N \rightarrow \infty} \sum_{n=1}^N \left( \int_X |\chi_{F_n}|^p dm \right)^{1/p} \\
\stackrel{\chi \in \{0,1\}}{=}\lim_{N \rightarrow \infty} \sum_{n=1}^N \left( \int_X \chi_{F_n} dm \right)^{1/p} \\
= \lim_{N \rightarrow \infty} \sum_n^N \left( m(F_n) \right)^{1/p} \\
\leq \lim_{N \rightarrow \infty} \sum_n^N \left( e^{-n} \right)^{1/p} < \infty
\end{align*}$$
Where the last inequality comes from the fact that this is a geometric series with $\frac{1}{e^{\frac{n}{p}}} < 1$ for $n$ large enough so it converges.
Edit
For the second function note that there exists an $N_0$ such that for $n > N_0: \Big | \frac{n^t}{e^n} \Big | < 1$. Then $$ \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{n^t}{e^n} = \sum_{n=1}^{N_0} \frac{n^t}{e^n} + \lim_{N \rightarrow \infty} \sum_{n={N_0}}^N \frac{n^t}{e^n} $$