For the normal distribution, if the pdf of $X$ is $f(x)=\frac{1}{\sqrt{32 \pi}} \exp[-\frac{(x+7)^2}{32}]$, $-\infty<x<\infty$, then $X$ is $N(-7,16)$ with a mean $\mu=-7$, a variance $\sigma^2=16$, and the moment generating function $M(t)=\exp(-7t+8t^2)$.
Question 1: I don't see why has normal distribution of $N(-7,16)$, and how to explain the moment generating function $M(t)=\exp(-7t+8t^2)$?
In general, if $X \sim N(\mu,\sigma^2)$, then it has pdf $$ f_{\mu,\sigma^2}(x) = \tfrac1{\sqrt{2\pi\sigma^2}} \exp\bigl( - \tfrac1{2\sigma^2} ( x - \mu )^2 \bigr) \quad\text{for}\quad x \in \mathbb R. $$ This is just the definition of a normal distribution. Setting $\mu := -7$ and $\sigma^2 := 16$, one recovers the special case.
Now, to find the moment generating function, we need to calculate $$ \textstyle M(t) = \mathbb E(e^{tX}) = \int_{-\infty}^\infty e^{tx} f_{\mu,\sigma^2}(x) dx = \tfrac1{\sqrt{2\pi\sigma^2}} \int_{-\infty}^\infty e^{tx} \exp\bigl( - \tfrac1{2\sigma^2} ( x - \mu )^2 \bigr) dx. $$ Can you try to solve this?
Hint: let $y := (x - \mu)/\sigma$; this is a "renormalised" version. Note that $(X - \mu)/\sigma \sim N(0,1)$. With this change of variables, you need only calculate it for $N(0,1)$, ie $\mu = 0$ and $\sigma^2 = 1$. To solve the integral, combine the two $e^{...}$ terms and complete the square.