We have to check convergence almost surely of $X_n$ to $0$
$$P(X_n=e^n)=\frac{1}{n+1}, \quad P(X_n=0)=1-\frac{1}{n+1}$$
I'm not sure I understand in practice the convergence Almost Surely and its difference from convergence in Probability.
For 2, here are two realizations which give different answers:
Assume in addition that $(X_n)$ are independent. Then by the 2nd Borel-Cantelli's lemma together with the estimate $$ \sum_{n=0}^{\infty} \Bbb{P}(X_n = e^n) = \infty, $$ it follows that $\Bbb{P}(X_n = e^n \text{ i.o.}) = 1$.Therefore in this case, $(X_n)$ diverges almost surely.
Now assume that $(X_n)$ is constructed in the following way: $(Z_n)_{n\geq1}$ are independent and $$ \Bbb{P}(Z_n = e) = \frac{n}{n+1}, \qquad \Bbb{P}(Z_n = 0) = \frac{1}{n+1}. $$ Now let $X_0 = 0$ and $X_n = Z_1 \cdots Z_n$. Then $$ \Bbb{P}(X_n = e^n) = \prod_{i=1}^{n} \Bbb{P}(Z_i = e) = \prod_{i=1}^{n}\frac{i}{i+1} = \frac{1}{n+1}, \qquad \Bbb{P}(X_n = 0) = 1 - \frac{1}{n+1}. $$ Notice that $\Bbb{P}(Z_n \neq 0 \text{ for all }n) = \prod_{n=1}^{\infty}\frac{n}{n+1} = 0$. Thus $Z_n = 0$ for some $n$ almost surely, which means that $X_n \to 0$ almost surely.
This does not converges Almost Surely, though it converges in probability.
For as converge to hold (in rather informal terms) we need that "almost all realizations" of the sequence $\{X_n \}$ tend to zero. In our case, an individual sequence tends to zero only if there exists some $M$ such that $X_n=0$ for all $n\ge M$. Then, we need to compute this probability.
Let's call $A$ the set of such sequences, and for each $\{X_n \}\in A$ define $M$ as the greatest index such that $X_M\ne 0$. Lets call $A_M$ the sets grouped by this value, so $A= \cup_{M=0}^\infty A_M$
Then $$P(\{X_n\} \in A_M)= P(X_M\ne 0 \wedge X_{M+1}=0 \wedge X_{M+2}=0 \dots)=\frac{1}{M+1} \prod_{k=1}^ {\infty} \frac{M+k}{M+k+1} =0$$
This implies that $P(\{X_n\} \in A) =0$ . But for as convergence we wanted $P(\{X_n\} \in A) =1$