Hint: Compute $E[S_T]$ in two ways.
On the one hand, $T$ is a stopping time, and $S$ is martingale, so $E[S_T]=E[S_0]$.
On the other hand, $S_T$ is either equal to $-3$ or $7$, so we have the simple expression $E[S_T]=(-3)\cdot P(S_T=-3)+7\cdot P(S_T=7)$.
Combining these observations lets you solve for $P(S_T=-3)$.
To prove that $T$ is a stopping time, you just need to show that the event $\{T\le k\}$ is contained in $\sigma(X_1,X_2,\dots,X_k)$. This is clear, since by looking at the first $k$ values of the sequence $X_1,\dots,X_k$, you can determine whether or not the process has hit $-3$ or $7$ by that point.
Proving $E[T]<\infty$ is a bit more involved. I gave a sketch of the proof in this answer.
To show that $P(T<\infty)=1$, consider the stopped martingale $\tilde X_n=X_{\min(n, T)}$. Since $\tilde X_n$ is a bounded martingale, the martingale convergence theorem implies that the sequence $\tilde X_n$ converges with probability one. But the only way an integer valued sequence can converge is if it is eventually constant, which implies that $T$ is finite.
Indeed, the fact that we cannot use the martingale convergence theorem is because $\sup_{n\geqslant 1}\mathbb E\left[\left\lvert Y_n\right\rvert\right]$ is infinite. Indeed, denoting $A_n=\{Y_{n-1}=0\}$ and $B_n=\{Y_{n-1}\neq 0\}$ gives
$$
\left\lvert Y_n\right\vert=\left\lvert X_{n}\right\vert\mathbf{1}_{A_{n-1}}+n\left\lvert Y_{n-1}\right\vert\left\lvert X_n\right\vert\mathbf{1}_{B_{n-1}}
$$
and since $X_n$ is independent of $\mathcal F_{n-1}=\sigma(X_1,\dots,X_{n-1})$ and $Y_{n-1}$ is $\mathcal F_{n-1}$-measurable, we derive that
$$
\mathbb E\left[\left\lvert Y_n\right\vert\right]
=\mathbb E\left[\left\lvert X_n\right\vert\right]\mathbb P\left(A_{n-1}\right)
+n\mathbb E\left[\left\lvert Y_{n-1}\right\vert\right]\mathbb E\left[\left\lvert X_n\right\vert\right]\
$$
and since $\mathbb E\left[\left\lvert X_n\right\vert\right]=1/n$, writting $p_n=\mathbb P(A_n)$, we get
$$
\mathbb E\left[\left\lvert Y_n\right\vert\right]=\frac{p_{n-1}}n+\mathbb E\left[\left\lvert Y_{n-1}\right\vert\right] .
$$
Moreover,
$$
p_n=\mathbb P(Y_n=0,Y_{n-1}=0)+\mathbb P(Y_n=0,Y_{n-1}\neq 0)
=\mathbb P(X_n=0,Y_{n-1}=0)+\mathbb P(X_n=0,Y_{n-1}\neq 0)=1-1/n
$$
hence letting $A_n=\mathbb E\left[\left\lvert Y_n\right\vert\right]$, we got
$$
A_n=\frac 1n-\frac 1{n(n-1)}+A_{n-1}
$$
and writing
$$
A_N-A_1=\sum_{n=2}^N\left(A_n-A_{n-1}\right)=\sum_{n=2}^N\left(\frac 1n-\frac 1{n(n-1)}\right)
$$
combined with the divergence of $\sum_{n\geqslant 2}1/n$ and the convergence of $\sum_{n\geqslant 2}1/(n(n-1))$ shows that $(A_N)_{N\geqslant 1}$ is not bounded.
Best Answer
Note that $Y_{n+1}=Y_n e^{X_{n+1}}$ so $$ E(Y_{n+1}\mid \mathcal{F_n})=Y_{n}E(e^{X_{n+1}}\mid \mathcal{F}_n)=Y_nEe^{X_{n+1}}\geq Y_n\exp(EX_{n+1})=Y_n $$ where we used the fact that $Y_n$ is measurable w.r.t to $\mathcal{F_n}$, the fact that $X_{n+1}$ is independent of $\mathcal{F_n}$ and Jensen's inequality in the final step. So $Y_n$ is a sub-martingale and not a martingale in general (see the equality conditions of Jensen's inequality for when it is a martingale)