Moment generating function questions with two distributions 
Suppose $X_1,X_2,\dots,X_n$ are independent each with exponential distribution with parameter $\lambda$ and $N$ has a Poisson distribution ($\lambda$) and is independent of $X_is$
a) Determine the moment generating function $S_N= X_1 + X_2+\dots+X_N$.
b) Use your answer in (a) to find mean of $S_N$.
c) Find the distribution $Y= \sum_{i=1}^n 2\lambda X_i$ identify the distribution of $Y$ by name and specify all parameter values.

I can only suspect this question has a lot to do with Gamma distribution's moment generating function. Need serious help on this kind of question, any help is really appreciated

As pedrosuavo suggests, the key here is conditional expectation; and, more specifically, the formula 
$$\tag{1}
\DeclareMathOperator{\E}{\mathbb{E}}\E\bigl[\E[e^{tS_N}\mid N]\bigr]=\E[e^{tS_N}].
$$
Conitioned on $N=n$, $S_N$ is distributed as $S_n$, so that
$$
\E[e^{tS_N}\mid N=n]=\E[e^{tS_n}\mid N=n]=\E[e^{tS_n}],
$$
since $S_n$ and $N$ are independent. But, since $X_1,\ldots,X_n$ are independent and identically distributed, we have
$$
\E[e^{tS_n}]=\E[e^{tX_1}\cdots e^{tX_n}]=\prod_{i=1}^{n}\E[e^{tX_i}]=(\E[e^{tX_1}])^n
$$
provided $n\geq 1$. This being the case, we know that
$$
\E[e^{tX_1}]=\int_0^{\infty}\lambda e^{-\lambda x}e^{tx}\,dx.
$$
Thus the function $t\mapsto \E[e^{tX_1}]$ has domain $t\in(-\infty,\lambda)$, and for such $t$ we have $\E[e^{tX_1}]=\frac{\lambda}{\lambda-t}$.
So, for $n\geq 1$, and $t<\lambda$, we have
$$
\E[e^{tS_N}\mid N=n]=\left(\frac{\lambda}{\lambda-t}\right)^n.
$$
What about when $n=0$? We have $S_0=0$, so that $\E[e^{tS_0}]=1=\left(\frac{\lambda}{\lambda-t}\right)^0$.
So, going back to the formula (1), we have that for $t<\lambda$,
$$
\begin{align}
\E[e^{tS_N}]&=\sum_{n=0}^{\infty}\E[e^{tS_N}\mid N=n]\cdot P(N=n)\\
&=\sum_{n=0}^{\infty}\left(\frac{\lambda}{\lambda-t}\right)^n\frac{e^{-\lambda}\lambda^n}{n!}\\
&=\exp\left[\frac{t\lambda}{\lambda-t}\right].
\end{align}
$$
Based on this, we can compute $\E[S_N]$; if $\phi(x)$ is the MGF for $S_N$, then we know that $\E[S_N]=\phi'(0)=1$.
(Conveniently, this matches our heuristic idea that $\E[S_N]$ should be something like $\E[N]\cdot\E[X_i]=\lambda\cdot\lambda^{-1}$.)
