Solution:

(i) We can write the given question as,
$\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{\sqrt{2}}=\sin ^{-1} \frac{1}{2}-\sin ^{-1}\left(2 \times \frac{1}{\sqrt{2}} \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}\right)$
On simplifying, we obtain
$=\sin ^{-1} \frac{1}{2}-\sin ^{-1}(1)$
By substituting the corresponding values, we obtain
$\begin{array}{l}
=\frac{\pi}{6}-\frac{\pi}{2} \\
=-\frac{\pi}{3}
\end{array}$

(ii) We can write the given question as
It is known that $\left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)=\pi / 3$
$=\sin ^{-1}\left\{\cos \left(\frac{\pi}{3}\right)\right\}$
Now substituting the values we obtain,
$\begin{array}{l}
=\sin ^{-1}\left\{\frac{1}{2}\right\} \\
=\frac{\pi}{6}
\end{array}$