What am I doing wrong? Joint Density of sum of random variables! So, I have this question:

And, what I have tried so far is I let $z_1 = \frac{X + Y}{2}$ and let $z_2 = \frac{X - Y}{2}$ to use change of variables for this question. However, when I work this out and plug it into my joint distribution of $X$ and $Y$ above, I end up with the $z_2$ disappearing and only having $z_1$ left in my joint distribution of $z_1$ and $z_2$. So, then when I go to find the marginal for just $z_1$ (which is $\frac{X + Y}{2}$ <- what I am looking for), there is no $z_2$ therefore, integrating wrt to $z_2$, I get $z_2$ just come up (integral of 1). This is an issue because the bounds are from 0 to $\inf$. Obviously, I can't have infinite in my marginal density so what am I doing wrong? 
Also, for question 5, I work it out similarly except I let $z_2 = X$ but I just end up with the exact same density as $f_{x,y}$ as seen in question 4 - is that normal? Please help, I been doing this question for a long time trying various dummy transformations.
Also, for question 4, I found the marginals of $x$ and $y$ and realized that the independence consequence (of the product being equal to the joint) is satisfied but using convolution, I get the same issue of my integral coming out to infinite! Thank you for all the help! 
EDIT: Am I doing my bounds wrong? the bound for z_2 is from 0 to inf or 0 to z_1?
 A: First, you can see that $X$ and $Y$ are independent. This is because the joint pdf is: $f_{XY}(x,y)=e^{-(x+y)}=e^{-x} e^{-y}$ (read this for independent random variables if you are not sure).
Also, you can get pdf of $X, Y$ by integration of the joint pdf.
$$f(x)=\int_0^{\infty}e^{-(x+y)}dy=e^{-x}\\f(y)=\int_0^{\infty}e^{-(x+y)}dx=e^{-y}$$
After you know the independence of $X,Y$, to find the distribution $(X+Y)/2$ will be not difficult.
I would like to use moment generating functions for this kind of problems.
Let $Z=\frac{X+Y}{2}$
$$E(e^{tZ})=E[e^{\left( t\frac{X+Y}{2} \right)}]=E(e^{t\frac{X}{2}} e^{t\frac{Y}{2}})$$
$\text{(by independence of X, Y )}$
$$=E(e^{t\frac{X}{2}})E(e^{t\frac{Y}{2}})=\int_{0}^{\infty}e^{t\frac{x}{2}}e^{-x}dx \times \int_{0}^{\infty}e^{t\frac{y}{2}}e^{-y}dy=\frac{1}{(1-\frac{t}{2})^2} , \text{t<2 here}$$
i.e. $$E(e^{tZ})=\frac{1}{(1-\frac{t}{2})^2}$$
$\therefore$ $Z$ has a gamma distribution with $\beta=1/2$ and $\alpha=2$
i.e $f(z)=\frac{1}{\Gamma(1/2)(1/2)^2}ze^{-2z}, 0<z<\infty$
For questions 5. You can use joint distribution of $f_{XY}(x,y)$ and Jacobian, such as let $Z =X+Y$ then $Y=X-Z$. It should not be very difficult
