Answer
$\color{blue}{P(\frac{1}{4}\le Y\le \frac{3}{4}) =\displaystyle \int^{3/4}_{1/4} 12(y^2-y^3)\ dy = \frac{11}{16}}$
Work Step by Step
$\underline{\text{Find}\ f_Y(y)}$
Writing $F_Y(y)$ as a split function and taking its first derivative with respect to $y$, where it is differentiable gives:
$\begin{align*}
F_Y(y) &= \begin{cases} 0, & y \lt 0 \\ 4y^3-3y^4, & 0\le y\le 1 \\ 1,& y \gt 1 \end{cases} \\
\frac{dF_Y(y)}{dy} &= \begin{cases} 0, & y \lt 0 \\ 12y^2-12y^3, & 0\lt y\lt 1 \\ 0,& y \gt 1 \end{cases} \\
f_Y(y) &= \begin{cases} 0, & y \lt 0 \\ 12(y^2-y^3), & 0\lt y\lt 1 \\ 0,& y \gt 1 \end{cases} \\
\color{blue}{f_Y(y)} &\color{blue}{= 12(y^2-y^3),\ 0 \lt y\lt 1}.
\end{align*}$
$\underline{\text{Using}\ f_Y(y), \text{find}\ P(\frac{1}{4}\le Y\le \frac{3}{4})}$
$\begin{align*}
P({\scriptsize \frac{1}{4}}\le Y\le {\scriptsize \frac{3}{4}}) &= \int^{3/4}_{1/4} f_Y(y)\ dy \\
&= \int^{3/4}_{1/4} 12(y^2-y^3)\ dy \\
&= 12\left(\frac{y^3}{3} - \frac{y^4}{4}\right]_{1/4}^{3/4} \\
&= \left( 4y^3 - 3y^4 \right]_{1/4}^{3/4} \\
&= 4\left((3/4)^3 - (1/4)^3\right) -3((3/4)^4 - (1/4)^4) \\
&= 4\left( \frac{27-1}{4^3}\right) - 3\left( \frac{81-1}{4^4}\right) \\
&= \frac{26}{4^2} - 3\cdot \frac{80}{4^4} \\
&= \frac{26}{4^2} - \frac{15}{4^2} \\
\color{blue}{P({\scriptsize \frac{1}{4}}\le Y\le {\scriptsize \frac{3}{4}})} &\color{blue}{= \frac{11}{16}}
\end{align*}$
