How to compute a marginal probability function from a joint probability function? 
I am looking at part a) and I have found the marginal p.f for $Y$ to be $e^{-2}2^{y}/y!$. I have set up for the equation for the marginal p.f for $X$ but I have no idea how to start it. Help would be appreciated. 
 A: You can see right away that these random variables are not independent from their support, as @Dilip Sarwate noted. In general if the joint support is not a product space, you can immediately conclude that the RVs are not independent. This is not a necessary condition though, so be careful. 
This is a one-liner then but let's derive the marginal densities as well.
\begin{align} f_X (x)=\sum_{y=x}^{\infty} f_{X,Y} (x,y) & =\sum_{y=x}^{\infty} \frac{e^{-2}}{x! \left(y-x\right)!}\\ &= \frac{e^{-2}}{x!} \sum_{y=x}^{\infty} \frac{1}{\left(y-x\right)!} \\&= \frac{e^{-1}}{x!} \quad (\text{why?}) \end{align} 
The marginal support of $X$ consists now of all nonnegative integers, i.e. $S_X=\left\{0,1,\ldots\right\}$. For $Y$ we similarly sum over all possible $X$ values.
\begin{align} f_Y (y)=\sum_{x=0}^{y} f_{X,Y} (x,y) &= \sum_{x=0}^{y}\frac{e^{-2}}{x! \left(y-x\right)!} \\ &= \frac{e^{-2}}{y!} \sum_{x=0}^{y} \binom{y}{x} \\ &=\frac{e^{-2}2^y}{y!} \quad (\text{why?}) \end{align}
Likewise, $S_Y=\left\{0,1,\ldots \right\}$. This pmf is very standard, can you recognize its type? From the marginal distributions now, it is obvious that $f_X(x) \times f_Y (y) \neq f_{X,Y} \left(x,y \right)$, as we had initially suspected. 
What you should take out of this is a look at the joint support is a quick way to verify independence and that when summing/integrating over random variables to arrive at a marginal density it is very important to have all restrictions in place. As verification of your work, you can always check whether the resulting mass function/density sums/integrates to $1$.
A: When dealing with bivariate discrete distributions it always helps to write joint distribution as a table. In this case we have (I omit the multiplicative constant $e^{-2}$. Also note the empty cells where distribution is not defined)
\begin{array} {|r|r|r|r|r|r|...}
\hline
 &y=0 & y=1 & y=2 & y=3\\ 
\hline
x=0& \frac{1}{0!0!} & \frac{1}{0!1!} & \frac{1}{0!2!} & \frac{1}{0!3!} & ...\\
\hline
x=1&  & \frac{1}{1!0!} & \frac{1}{1!1!} & \frac{1}{1!2!} & ...\\
\hline
x=2& &  & \frac{1}{2!0!} & \frac{1}{0!1!} & ...\\
\hline
x=3&  &  &  & \frac{1}{3!0!} & ...\\
\hline
... & & & & & ...\\
\hline
\end{array}
Then marginal distributions are simply the sums of columns (for $Y$) or rows (for $X$).
Looking at the rows it should not be hard to spot the pattern and get the marginal distribution for $X$. For $Y$ it is a bit harder, since you need to recall the formula for $n \choose m$ and the fact that $\sum_{m=0}^n {n \choose m}=2^n$.
