b
($1$) $I=2 \int_0^{\frac{\pi}{2}} \underbrace{\sin ^2 x \sqrt{\frac{\pi x}{2}-x^2}}_{h_1}-\int_0^{\frac{\pi}{2}} g(x) d x$
Let $I_1=\int_0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$
(making perfect square)
apply kings
$I_1=\int_0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$
add both
$2 I_1=\int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$
i.e. $2 I_1=\int_0^{\frac{\pi}{2}} g(x)$
Now $I=2 I_1-\int_0^{\frac{\pi}{2}} g(x)=0$
($2$) Now $I_1=\int_0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) d x$
i.e. $\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} d x$
Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$
$\Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\pi^2}{\frac{16}{2}} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2}$
$\Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right]$
$\Rightarrow \frac{1}{2} \times \frac{\pi^3}{32}$
Now $\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$