If $M$ is a manifold and $\hat M$ its universal cover, $\hat M$ is endowed with a differentiable structure such that $p:\hat M\rightarrow M$ is differentiable. For every $x\in M$, there exists a neighborhood $U$ of $x$ such that $U$ is a domain chart $\phi:U\rightarrow\mathbb{R}^n$ and $p^{-1}(U)=\bigcup V_i$ and $p_{\mid V_i}\rightarrow U$ is an homeomorphism, the differentiable structure is defined by supposing that $p_i$ is a diffeomorphism which induces the chart $\phi\circ p_i$ on $\hat V_i$.
Now look at the action of an element $\gamma\in\pi_1(M)$ on $\hat M$, i.e. a covering transformation. We will prove that this action is already smooth. Let $p(y)=x$. There exist $i$ and $j$ such that $y\in V_i$ and $\gamma . y\in V_j$,
However, the map $(\phi\circ p_j) \circ \gamma \circ (\phi\circ p_i)^{-1}$ is the identity, this implies that action of $\gamma$ is differentiable.
This proves that every covering transformation is smooth.
This is a nice question. The following is an argument for the first part. I'll assume that $\tilde{X}$ is connected since you called it the universal cover.
Let $\tilde{Y}=p^{-1}(Y)$. Then, it is clear that $q=p \mid_{\tilde{Y}}$ is also a covering map. Moreover, $q_*$ is injective. Now, let $i_*:\pi_1(Y) \to \pi_1(X)$ be induced by inclusion, and let $\tilde{i}$ be its lift. Then there is an equality
$i*\circ q_*= p_* \circ \tilde{i}_* $,
but since $i_* \circ q_*$ is injective, we can deduce that $\tilde{i}_*$ is as well, so $\pi_1(\tilde{Y})=0$.
I've omitted basepoints everywhere but this works on each component of the preimage.
The previous argument can be strengthened if the inclusion map gives an isomorphism on the fundamental group.
If $(K,L)$ is a $CW$ pair such that $i_*:\pi_1(L) \to \pi_1(K)$ is an isomorphism, and $\tilde{K}$ is the universal cover of $K$, then we can strengthen the last argument to show that $\tilde{L}$ is connected. Indeed, we know that $p_i(\tilde{K},\tilde{L}) \cong \pi_i(K,L)$ for $i \geq 1$ by the homotopy lifting property, and by the LES of a pair, we have that
$$\pi_1(L) \to \pi_1(K) \to \pi_1(K,L) \to \pi_0(L) \to \pi_0(K)$$
is exact, but by our assumption on connectedness, the first and last maps are isomorphisms, from which we can deduce that $\pi_1(\tilde{K},\tilde{L})=0$, so applying this sequence again to the covers, we see that
$$0 \to \pi_0(\tilde{L}) \to \pi_0(\tilde{K}) $$
is exact (When interpreted correctly), so $\tilde{L}$ is connected.
The reason I went on my tangent in the middle was because I have a proof idea for (2) that I will try to flesh out, but I have a strong feeling that I'm being an idiot here.
We know that $i_*:\pi_1(Y,y_0) \subset \pi_1(X,x_0)$, so there is a corresponding (Connected) cover $r:X_1 \to X$ such that $\pi_1(X_1,x_1)=i_*(\pi_1(Y,y))$.
It can be shown that right cosets of $r_*(\pi_1(X_1,x_1))=i_*\pi_1(Y,y_0)$ are in bijection with lifts of $x_0$ in $X_1$ (see prop VI.19 here .)
Moreover, we know that there is a lift $\tilde{r}:(Y,y_0) \to (X_1,x_1)$ of $i:(Y,y_0) \to (X,x_0)$ that is an isomorphism on $\pi_1$ with the image connected, and hence $((X_1,x_1),r(Y,y_0))$ satisfies the conditions of the middle paragraph so the lift of $r(Y,y_0))$ is connected in $\tilde{X}$.
I hope that this can yield a proof with maybe some more effort. The idea is that we have reduced the problem to showing that preimages of basepoints in $X_1$ correspond to connected components of $\tilde{Y} \subset \tilde{X}$.
Best Answer
Yes it is. Let $X$ be a topological space with universal cover $\widetilde{X}$ and covering map $p : \widetilde{X} \to X$. A homeomorphism $f : \widetilde{X} \to \widetilde{X}$ is called a deck transformation of $p$ if $p\circ f = p$; that is, $f$ preserves the fibres of $p$ so if $y \in p^{-1}(x)$, $f(y) \in p^{-1}(x)$. The set of all deck transformations of $p$ is denoted $\operatorname{Deck}(p)$ and forms a group under composition. The quotient of $\widetilde{X}$ by $\operatorname{Deck}(p)$ is $X$. Moreover, $\operatorname{Deck}(p) \cong \pi_1(X)$.
A good reference for this material is Hatcher's Algebraic Topology.