Finding distribution function of $Y/X$ and probability density function of $X+Y$

There is a "lather rinse repeat" way of solving such problems. I'll show you some steps to do the first one, you can try the technique on the second one too.

\begin{eqnarray*} F_Z(z) &=& \Pr(Z \leq z) \\ &=& \Pr(Y/X \leq z) \\ &=& \int_0^\infty \Pr(Y \leq zx) f_X(x) dx \\ &=& \int_0^\infty (1-e^{-zx})xe^{-x} dx. \end{eqnarray*}

Run this integral through, and that should do it. Of course, some of these steps may require careful justifications: for instance going from $Y/X \leq z$ to $Y \leq zX$ requires care with the sign of $X$. I hope you can figure these details out.

Alternatively, for the second one, you may want to try going the moment generating function/characteristic function route.

It is so late here almost 2 am but I can explain the way to follow. First of all the two questions are related to *function of a random variable*$\rightarrow$ check in wikipedia.

In the second question you have $Z=X+Y$ and it corresponds to the convolution of two marginal pdfs. Simply you will convolve $f_X(x)$ with $f_Y(y)$ and you will get $f_Z(z)$.

In the first one can again use the concept of function of a random variable. Since the pdfs are defined in the positive real line, I guess it is easier to use the identity

$$\log(Z)=\log(Y)-\log(X)$$

From here again using function of a random variable one can calculate simply the pdfs of $\log(Y)$ and $-\log(X)$. Again the pdf of $\log(Z)$ will be the convolution of these two pdfs that you found. Now we have a random variable $\log(Z)$ with pdf $f_{\log(Z)(\log(z))}$ to get $f_Z(z)$ again we need to apply function of a random variable now our function is $\exp$ and the related pdf is the one which we found $f_{\log(Z)(\log(z))}$. The result will give $f_Z(z)$.

I hope it helps. Have a good night.