Moment Generating Function Questions on Technique 
In this first question, I can work out that $X_i\sim \operatorname{Exp}(\lambda)$ and $N\sim \operatorname{Geom}(p)$.
Do I need the MGF of $T = \sum X_i$? How can I calculate this and how would I use this to come up with the answer?

In this example, I can work out the first 2 parts which are standard properties of MGFs, but how do I go about calculating the second 2?
Any help would be greatly appreciated! Thank you 
 A: The reason that the MGF is useful, is because it characterizes the distribution of a r.v completely, that is, if you know the MGF then you know the distribution and vice versa. 
As you suggest, you have to compute the distribution of $X := \sum_{i=1}^{N} X_i$, where $N \sim \mbox{Geom}(p)$. Since the upper bound of the summation is a r.v as well, you will need conditioning here:
$$\mbox{MGF}_X(t) = \mathbb{E}[\exp(tX)] = \sum_{n \geq 1} \mathbb{E}\left[\exp\left( t \sum_{i=1}^{n} X_i\right) \ | \ N = n \right]\mathbb{P}[N=n].$$
To proceed, use that $N$ is independent of the duration times $X_i$ and the fact that $X_i$ are i.i.d.
A: For the second question:
(iii) $Z := UX$ has the required mgf:
\begin{align}
M_{Z}(t) &= E(e^{tUX}) \\
&= \int_{u=0}^{1}f_U(u)E(e^{tUX}\mid U=u)\;du \\
&= \int_{u=0}^{1}E(e^{tuX})\;du \\
&= \int_{u=0}^{1}M_X(tu)\;du.
\end{align}
(iv) Let:
$$Z =
\begin{cases}
X,  & \text{if $0\leq U\leq 1/2$} \\
Y,  & \text{if $1/2\lt U\leq 1$}.
\end{cases}$$
Then,
\begin{align}
M_Z(t) &= P(U\leq 1/2)E(e^{tZ}\mid U\leq 1/2) + P(U\gt 1/2)E(e^{tZ}\mid U\gt 1/2) \\
&= \dfrac{1}{2}E(e^{tX}) + \dfrac{1}{2}E(e^{tY}) \\
&= \left[M_X(t) + M_Y(t)\right]/2.
\end{align}
