= Solution
Put $\mu=\mathbb E[X]$ and $\sigma^2=\operatorname{Var}(X)$. Independence of $X$ and the <uniform distribution> variable $U$ gives
$$
\mathbb E[Y]=\mathbb E[X]\mathbb E[U]=\boxed{\frac\mu2}.
$$
Also $\mathbb E[U^2]=1/3$, so
$$
\begin{aligned}
\operatorname{Var}(Y)
&=\mathbb E[X^2]\mathbb E[U^2]-\mathbb E[Y]^2\\
&=\frac{\sigma^2+\mu^2}{3}-\frac{\mu^2}{4}.
\end{aligned}
$$
Therefore
$$
\boxed{\operatorname{Var}(Y)=\frac13\operatorname{Var}(X)+\frac1{12}\mathbb E[X]^2}.
$$
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