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.
Okay, let's first see why the first binary digit of $U$ is Bernoulli$(1/2)$. The first binary digit is $1$ if and only if $U \geq 1/2$, which has probability $1/2$, so we are done. For convenience, let $B_n$ denote the $n^{th}$ binary digit of $U$. Now, inductively assume that $B_1,\ldots,B_{n-1}$ are i.i.d. Bernoulli$(1/2)$. Then, look at the conditional probability $q_n:=\mathbb{P}(B_n=1\big|(B_1,\ldots,B_{n-1})=(b_1,\ldots,b_{n-1}))$, for a sequence $(b_1,\ldots,b_{n-1}) \in \{0,1\}^{n-1}$. Divide the interval $[0,1]$ into the diadic intervals of length $1/2^{n-1}$, and let these intervals be enumerated from left to right as $I_1,I_2,...I_{2^{n-1}}$. Now, what does the event $(B_1,\ldots,B_{n-1})=(b_1,\ldots,b_{n-1})$ say? It says that (is equal to the following event) $U$ must lie in exactly one of these diadic intervals, say $I_i$, where $i$ is a (complicated, but don't need to know) deterministic function of the deterministic binary sequence $(b_1,\ldots,b_{n-1})$. The way to find this interval is to follow a binary search algorithm, similar to the proof of the Heini-Borel theorem in real analysis.
Anyway, let $m_i$ be the midpoint of $I_i$. So,
$$q_n = \mathbb{P}(U > m_i|U\in I_i)~.$$ The above probability is obviously $1/2$. This shows that $B_n$ has a Bernoulli$(1/2)$ distribution, independent of $(B_1,\ldots,B_{n-1})$, and the induction is complete.
Best Answer
Let $n=1$ and suppose $P(X=0)=1/3, P(X=1)=2/3$. Suppose $P(Y=0|X=0)=1$ and $P(Y=0|X=1)=1/2$. Then $X+Y$ is uniform on $\{0,1,2\}$.
Edit after comment: Let $X$ be uniformly distributed on the even numbers $\{0,2,4\}$ and let $Y$ be uniformly distributed on $\{0,1\}$, with $X$ and $Y$ independent. Note that $X$ is not uniformly distributed on $\{0,1,\dots, 4\}$, but that $X+Y$ is uniformly distributed on $\{0,1,2,3,4,5\}$.