Answer
a) See explanation
b) $\color{blue}{\frac{(b-a)^2}{12}}$
Work Step by Step
a)
We show that the pdf of $Y$ is a uniform distribution on $[a,b]$, i.e., $f_Y(y) = \dfrac{1}{b-a},\ a \le y \le b.$
The cdf of $Y$ is,
$\begin{align*}
F_Y(y) &= P(Y \le y),\ a \le y \le b \\
&= P((b-a)U + a \le y) \\
&= P\left( U \le \frac{y-a}{b-a}\right),\ a \le y \le b \\
&= \int_a^{\large \frac{y-a}{b-a}} 1\ du \quad \left[\ \text{since}\ f_U(u) = 1,\ 0 \le u \le 1,\ \right] \\
&= u\ \biggr\vert_0^{\large\frac{y-a}{b-a}} \\
F_Y(y) &= \frac{y-a}{b-a},\ a \le y \le b.
\end{align*}$
Thus,
$\begin{align*}
f_Y(y) &= F'(y) \\
&= \frac{d}{dy}\left(\frac{y-a}{b-a}\right),\ a \lt y \lt b \\
f_Y(y) &= \frac{1}{b-a},\ a \lt y \lt b.
\end{align*}$
Since $P(Y=a)=P(Y=b)=0$, we can write
$f_Y(y) = \dfrac{1}{b-a},\ a\le y \le b,$
so that the pdf of $Y$ is that of a uniform distribution on $[a,b]$.
b)
$\begin{align*}
E(Y) &= \int_{\mathbb{R}} y \cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y)\ ] \\
&= \int_a^b y\cdot \frac{1}{b-a}\ dy \qquad \left[\ \text{since}\ f_Y(y)= \frac{1}{b-a},\ a\le y \le b\ \right] \\
&= \left(\frac{y^2}{2}\cdot\frac{1}{b-a}\ \right\vert_a^b \\
&= \frac{b^2-a^2}{2} \cdot \frac{1}{b-a} \\
&= \frac{(b-a)(b+a)}{2} \cdot \frac{1}{b-a} \\\
E(Y)&= \frac{b+a}{2} \\
\mu &= \frac{b+a}{2}
\end{align*}$
$\begin{align*}
E(Y^2) &= \int_{\mathbb{R}} y^2 \cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\
&= \int_a^b y^2\cdot \frac{1}{b-a}\ dy \qquad \left[\ \text{since}\ f_Y(y)= \frac{1}{b-a},\ a\le y \le b\ \right] \\
&= \left(\frac{y^3}{3}\cdot\frac{1}{b-a}\ \right\vert_a^b \\
&= \frac{b^3-a^3}{3} \cdot \frac{1}{b-a} \\
&= \frac{(b-a)(b^2+ab+b^2)}{3} \cdot \frac{1}{b-a} \\\
E(Y^2)&= \frac{b^2+ab+a^2}{3}
\end{align*}$
By Theorem 3.6.1, since $\mu= \dfrac{b+a}{2}$ exists and $E(Y^2) = \dfrac{b^2+ab+a^2}{3}$ is finite,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - \mu^2 \\
&= \frac{b^2+ab+a^2}{3} - \left(\frac{b+a}{2}\right)^2 \\
&= \frac{b^2+ab+a^2}{3} - \frac{b^2+2ab+a^2}{4} \\
&= \frac{(4b^2+4ab+4a^2)-(3b^2+6ab+3a^2)}{12} \\
&= \frac{b^2-2ab+a^2}{12} \\
\color{blue}{\text{Var}(Y)}\ &\color{blue}{= \frac{(b-a)^2}{12}}
\end{align*}$
