Problem: If $MX(t)=e^{−5(1−et)}$ , find $V(X)$.
Also find: $P(X=3)$
My first intuition is that this would follow the Poisson rule: $M(t)=e^{\lambda(e^t-1)}$
This would mean that $\lambda=E(X)=-5?$
I am not sure how to get $E(X^2)$ so that I can finish the equation $V(X)=E(X^2)-(E(X))^2$.
For the $P(X=3)$, it would appear that This is the MGF of the following discrete $rv$ is as follows:
$$X = \begin{cases} 1, & \text{if $x=-5$ } \\ 0, & \text{if $x=0$ } \end{cases} $$
So I don't know that I'm working through this part correctly either.
The MGF of the Poisson as you have written would imply that $\lambda = 5$. The MGF is exactly what it sounds like, a generating function for the moments. $M^{(k)}(0) = E(X^k)$. Thus $M''(0) - (M'(0))^2 = \text{var}(X)$. You can show that this implies that $\text{var}(X) = 5$.
To find $P(X = 3)$, it is better to use the distribution itself $$ P(X = x) = \frac{e^{-5}5^x}{x!}. $$