We need to show that $\left\{Y_n\right\}$ is uniformly integrable, given that $\left\{X_n\right\}$ and $\left\{Z_n\right\}$ are each uniformly integrable. To do this, there are two defining properties that need to be checked. In what follows, I assume that the random variables are all defined on the same probability space.
Boundedness: Firstly, we show that $\,\,\sup\limits_{n}\left\{\mathbb{E}\left[\left|Y_n\right|\right]\right\}{}<{}\infty$.
Observe that, since $\,\,X_n(\omega)\leq Y_n(\omega)\leq Z_n(\omega)\,\,$ pointwise, then
$$
\left|Y_n\right|\leq\max\left\{\left|X_n\right|,\left|Z_n\right|\right\}{}={}\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}{}+{}\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}.
$$
Consequently, for each $n$,
$$
\begin{eqnarray*}
\mathbb{E}\left[\left|Y_n\right|\right]&{}={}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}{}+{}\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}={}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}\right]{}+{}\mathbb{E}\left[\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}\leq{}&\mathbb{E}\left[\left|X_n\right|\right]{}+{}\mathbb{E}\left[\left|Z_n\right|\right]\newline
&&\newline
&{}\leq{}&\sup\limits_{n}\left\{\mathbb{E}\left[\left|X_n\right|\right]\right\}{}+{}\sup\limits_{n}\left\{\mathbb{E}\left[\left|Z_n\right|\right]\right\}{}<{}\infty\,.
\end{eqnarray*}
$$
But, this is true for all $n$. Therefore, $\,\,\sup\limits_{n}\left\{\mathbb{E}\left[\left|Y_n\right|\right]\right\}{}<{}\infty$.
Arbitrarily Small Integrals: Choose $\epsilon>0$. There exist $\delta_x>0$ and $\delta_z>0$ such that, for all $n$ and events $A$ with $P\left(A\right)<\delta_y:=\min\left\{\delta_x,\delta_z\right\}$, we have
$$
\mathbb{E}\left[{\bf{1}}_{A}\left|X_n\right|\right]{}<{}\dfrac{\epsilon}{2}\,\,\,\,\mbox{ and }\,\,\,\,\mathbb{E}\left[{\bf{1}}_{A}\left|Z_n\right|\right]{}<{}\dfrac{\epsilon}{2}\,.
$$
So, for any $n$, under these circumstances we have,
$$
\begin{eqnarray*}
\mathbb{E}\left[{\bf{1}}_{A}\left|Y_n\right|\right]&{}\leq{}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{A}{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}\right]{}+{}\mathbb{E}\left[\left|Z_n\right|{\bf{1}}_{A}{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}\leq{}&\mathbb{E}\left[{\bf{1}}_{A}\left|X_n\right|\right]{}+{}\mathbb{E}\left[{\bf{1}}_{A}\left|Z_n\right|\right]\newline
&&\newline
&{}<{}&\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon\,.
\end{eqnarray*}
$$
Edit:@Kolmo points out, and I agree, that uniform integrability is not needed here. In fact, since $X_n, Y_n$ and $Z_n$ each converge in probability, they also converge in distribution. Consequently, the only thing we need to show is that
$$
\mathbb{E}\left[Y\right]<\infty.
$$
But, this must be the case, since $X$ and $Z$ satisfy
$$
\lim\limits_{n\to\infty}\mathbb{E}\left[X_n\right]{}={}\mathbb{E}\left[X\right]{}\leq{}\mathbb{E}\left[\left|X\right|\right]<\infty
$$
and
$$
\lim\limits_{n\to\infty}\mathbb{E}\left[Z_n\right]{}={}\mathbb{E}\left[Z\right]{}\leq{}\mathbb{E}\left[\left|Z\right|\right]<\infty,
$$
so, using the bound on $\left|Y_n\right|$ given above,
$$
\mathbb{E}\left[Y\right]{}={}\lim\limits_{n\to\infty}\mathbb{E}\left[Y_n\right]{}\leq{}\lim\limits_{n\to\infty}\mathbb{E}\left[\left|Y_n\right|\right]{}\leq{}\mathbb{E}\left[\left|X\right|\right]{}+{}\mathbb{E}\left[\left|Z\right|\right]{}<{}\infty\,.
$$
This is a direct consequence of Scheffé's lemma, which is actually due to Riesz:
Lemma: If a sequence of $L^p$ integrable functions $f_n$ converges a.e. to an $L^p$ integrable function $f$ with $p\geq 1$ and $\lim_n \Vert f_n \Vert_p = \Vert f \Vert_p$ holds true, then $\lim_n \Vert f_n - f \Vert_p = 0$.
Proof for p=1 (taken from Kusolitsch (2010)): Consider functions
$$ f_n^* = \begin{cases} f_n, & \vert f_n \vert \leq \vert f \vert, \\ \vert f \vert sgn(f_n), & \vert f_n \vert > \vert f \vert, \end{cases}$$
which are dominated by the $L^1$ integrable $\vert f \vert$ and converge to $f$ a.e. So the functions $\vert f_n^* - f \vert$ are dominated by $2\vert f \vert$ and vanish a.e., and the dominated convergence theorem yields
$$\lim_n \int \vert f_n^* \vert = \int \vert f \vert$$
and also
$$\lim_n \int \vert f_n^* - f \vert = 0.$$
By definition $f_n^*$ always has the same sign as $f_n$ and $\vert f_n^* \vert \leq \vert f_n \vert$. So one gets $\vert f_n - f_n^* \vert = \vert f_n \vert - \vert f_n^* \vert$, and
$$ \int \vert f_n - f_n^* \vert = \int \vert f_n \vert - \int \vert f_n^* \vert. $$
Since both integrals on the right hand side converge to $\int \vert f\vert<\infty$, this yields the conclusion.
Best Answer
Your first question seems oddly phrased as there is no connection between the $X_n$ and the $Y_n$ (except that they have the same almost sure limit - but this is very weak).
In any case, the answer is no. Work with $[0,1]$ with Borel sets and Lebesgue measure. Put $X_n:= n^2 \mathbb{1}_{[0,1/n]}$. Then $X_n \rightarrow 0$ a.s., but $EX_n = n$, so $(X_n)$ is not bounded in $L^1$. It consequently isn't UI. Put $Y_n := 0$ identically. Then $(Y_n)$ converges a.e. to $0$ and is UI.
Do you see what I mean by there being no connection?