$$
f(x)=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2
$$

Let
$$
\sin ^{-1} x=a \text { and } \cos ^{-1} x=b
$$

Then,
$$
\begin{aligned}
f(x) & =a^2+b^2 \\
& =(a+b)^2-2 a b
\end{aligned}
$$

Put the value of $a$ and $b$
$$
\begin{aligned}
& f(x)=\left(\sin ^{-1} x+\cos ^{-1} x\right)^2-2 \sin ^{-1} x \cos ^{-1} x \\
& =\frac{\pi^2}{4}-2 \sin ^{-1} x \cos ^{-1} x\left[\because \sin ^{-1} x+\cos ^{-1} x=\pi / 2\right] \\
& \quad=\frac{\pi^2}{4}-2 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\
& =\frac{\pi^2}{4}-\pi \sin ^{-1} x+2\left(\sin ^{-1} x\right)^2
\end{aligned}
$$
For minimum and maximum value,
$$
\begin{aligned}
f^{\prime}(x) & =0-\pi \cdot \frac{1}{\sqrt{1-x^2}}+4 \sin ^{-1} x \frac{1}{\sqrt{1-x^2}}=0 \\
& =\frac{1}{\sqrt{1-x^2}}\left[4 \sin ^{-1} x-\pi\right]=0=\sin ^{-1} x=\pi / 4 \\
x & =\sin \pi / 4=\frac{1}{\sqrt{2}}
\end{aligned}
$$

Therefore, $f^{\prime \prime}\left(\frac{1}{\sqrt{2}}\right)=+$ ve
$$
\begin{aligned}
& f(x)_{\min }=\text { when } x=1 / \sqrt{2} \\
& \therefore \quad f(x)_{\min }=\frac{\pi^2}{4}-2 \sin ^{-1} x \cos ^{-1} x \\
& \therefore \quad f(x)_{\min }=\frac{\pi^2}{4}-2 \cdot \frac{\pi}{4} \cdot \frac{\pi}{4} \\
& \alpha=\frac{\pi^2}{8} \\
& f(x)_{\max }=\frac{\pi^2}{4}-2 \sin ^{-1} x \cos ^{-1} x \\
&
\end{aligned}
$$
We can see that, $f(x)$ is maximum, when $x=-1$
$$
\begin{gathered}
f(x)_{\max }=\frac{\pi^2}{4}-2\left(-\frac{\pi}{2}\right)(\pi) \\
\beta=\frac{\pi^2}{4}+\pi^2=\frac{5 \pi^2}{4} \Rightarrow \beta=\frac{5 \pi^2}{4} \\
\text { Hence, } 8(\alpha+\beta)=8\left[\frac{\pi^2}{8}+\frac{5 \pi^2}{4}\right]=8\left[\frac{\pi^2+10 \pi^2}{8}\right] \\
=11 \pi^2
\end{gathered}
$$