Answer :
a) [tex]Z=\frac XY[/tex] has CDF
[tex]F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy[/tex]
[tex]F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy[/tex]
where the last equality follows from independence of [tex]X,Y[/tex]. In terms of the distribution and density functions of [tex]X,Y[/tex], this is
[tex]F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy[/tex]
Then the density is obtained by differentiating with respect to [tex]z[/tex],
[tex]f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy[/tex]
b) [tex]Z=XY[/tex] can be computed in the same way; it has CDF
[tex]F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy[/tex]
[tex]F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy[/tex]
Differentiating gives the associated PDF,
[tex]f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy[/tex]
Assuming [tex]X\sim\mathrm{Exp}(\lambda_x)[/tex] and [tex]Y\sim\mathrm{Exp}(\lambda_y)[/tex], we have
[tex]f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy[/tex]
[tex]\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}[/tex]
and
[tex]f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy[/tex]
[tex]\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy[/tex]
I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.