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.
There is an error in the question. You have to replace $\frac {\lambda S_n-n\lambda^{2}} 2$ in the exponent defining $Z_n$ by $\lambda S_n-\frac {n\lambda^{2}} 2$. As it stands, $(Z_n)$ is not a martingale.
$\Bbb{E}\left(e^{\lambda X_{1}-\lambda^2/2 }\right)=e^{-\lambda^{2}/2} \Bbb{E} \left(e^{\lambda X_{1}}\right)$ and $\Bbb{E}\left(e^{\lambda X_{1}}\right)=e^{\lambda^{2}/2}$.
For the last equaion refer to MGF of normal distribution in Wikipedia.
Best Answer
Note that $X_{n+1}$ is independent of $\mathscr{F}_n$, while $S_n$ is $\mathscr{F}_n$-measurable. It follows that for any measurable (and admissible) $f$, we have $E[f(S_n,X_{n+1})|\mathscr{F}_n]=E[f(s,X_{n+1})]|_{s=S_n}$. Then $$\begin{aligned}E[P(S_{n+1},n+1)|\mathscr{F}_{n}]&=E[P(S_{n}+X_{n+1},n+1)|\mathscr{F}_{n}]=\\ &=\frac{1}{2}P(S_{n}+1,n+1)+\\ &+\frac{1}{2}P(S_{n}-1,n+1)=\\ &=\frac{1}{2}2P(S_n,n)=P(S_n,n)\end{aligned}$$