Answer
$\color{blue}{5/16}$
Work Step by Step
$\underline{\text{Using cdf}}$
$\begin{align*}
P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}}) &= P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}}) + 0 \\
&= P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}}) + P(Y={\scriptsize\frac{3}{4}}) \\
&= P({\scriptsize \frac{1}{2}} \lt Y\le {\scriptsize\frac{3}{4}}) \\
&= F_Y({\scriptsize\frac{3}{4}}) - F_Y({\scriptsize\frac{1}{2}}) \\
&= (3/4)^2 - (1/2)^2 \\
&= 9/16 - 1/4 \\
&= (9-4)/16 \\
\color{blue}{P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}})} &\color{blue}{= 5/16}
\end{align*}$
$\underline{\text{Using pdf}}$
$\begin{align*}
P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}}) &= \int_{1/2}^{3/4} f_Y(y)\ dy \\
&= \int_{1/2}^{3/4} \frac{dF_Y(y)}{dy}\ dy \\
&= \int_{1/2}^{3/4} \frac{d}{dy}y^2\ dy \\
&= \int_{1/2}^{3/4} 2y\ dy \\
&= \left. 2\cdot \dfrac{y^2}{2}\right]_{1/2}^{3/4} \\
&= \left. y^2\right]_{1/2}^{3/4} \\
&= ((3/4)^2 - (1/2)^2) \\
&= 9/16 - 1/4 \\
&= (9-4)/16 \\
\color{blue}{P({\scriptsize \frac{1}{2}} \lt Y\lt {\scriptsize\frac{3}{4}})} &\color{blue}{= 5/16}\end{align*}$
