In the first part of the question we showed that $P(X \geq k)\leq E(e^{tX}e^{-kt})$ for all $t \geq 0$ and real $k$ by the use of Markov's inequality. This wasn't too bad.
Now, in the second part, first we have to calculate $E(e^{tX})$ where $X$ is a Poisson random variable with parameter 1 and then use the above inequality with a suitable choice of $t$ to get $\frac{1}{k!}+\frac{1}{(k + 1)!}+\frac{1}{ (k + 2)! }+...\leq (\frac{e}{k})^k$ for $k\geq 1$.
If I am right then $E(e^{tX})=e^{e^{t}-1}$ but I couldn't show the last part.
$$\sum_{n\geq k}\frac{1}{n!}=\frac{1}{k!}\left(1+\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}+\ldots\right)\leq \frac{1}{k!}\sum_{j\geq 0}\frac{1}{(k+1)^j}=\frac{k+1}{k\cdot k!}$$ hence we just need to prove that: $$ \left(1+\frac{1}{k}\right)\frac{k^k}{k!}\leq e^k. \tag{1} $$ The inequality triviall holds for $k=1$; moreover, by setting $A_k = \left(1+\frac{1}{k}\right)\frac{k^k}{k!}$ we have: $$ \frac{A_{k+1}}{A_k}=\frac{k(k+2)}{(k+1)^2}\left(1+\frac{1}{k}\right)^k \leq \left(1+\frac{1}{k}\right)^k \leq e\tag{2},$$ hence $(1)$ holds for every $k\geq 1$ by induction.