We have $Z\le z$ if $Y/X\le z$, i.e. $Y\le zX$. For $z\le1$, the probability for this, given $X=x$, is $zx$, and then integrating over $x$ yields
$$
\int_0^1zx\,\mathrm dx=\frac z2\;.
$$
To get the result for $z\gt1$, consider $1/Z=X/Y$ and use the result for $z\le1$.
Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$,
$$
[y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z],
$$
hence, for every $0\lt y\lt z\lt1$,
$$
P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n.
$$
In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$,
$$
F_Z(z)=P(Z\leqslant z)=z^n.
$$
Thus, for every $(y,z)$ in $(0,1)$,
$$
F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z)
$$
is also
$$
F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}.
$$
One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as
$$
f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}.
$$
In the present case, for every $n\geqslant2$,
$$
f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}.
$$
Best Answer
On 1)
Let $W:=X+Y$. Then:
$$F_{W}\left(w\right)=P\left(X+Y\leq w\mid X=0\right)P\left(X=0\right)+P\left(X+Y\leq w\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}F_{Y}\left(w\right)+\frac{1}{2}F_{Y}\left(w-1\right)$$
Here $F_{Y}$ is well known to you and knowing CDF $F_{W}$ you can find PDF $f_{W}$.
On 2)
$X=0\Rightarrow XY=0$ so that $P\left\{ XY=0\right\} \geq P\left\{ X=0\right\} \geq\frac{1}{2}$. Draw your conclusions about the existence of a PDF.
On 3)
Let $V:=XY$. Then:
$$F_{V}\left(v\right)=P\left(XY\leq v\mid X=0\right)P\left(X=0\right)+P\left(XY\leq v\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}P\left(0\leq v\right)+\frac{1}{2}F_{Y}\left(v\right)$$
Here $P\left(0\leq v\right)=0$ if $v<0$ and $P\left(0\leq v\right)=1$ otherwise.