Ans : $\large \frac{2\pi}{5}$
$\large\frac{3\pi}{5}$ is not in the principal interval of sin.
Therefore, reduce it to its principal interval
$\large\frac{3\pi}{5}= \pi - \large\frac{2\pi}{5}$
$\sin\large\frac{3\pi}{5} =\sin(\pi- \large\frac{2\pi}{5})=\sin\large\frac{2\pi}{5} $
$\large\frac{2\pi}{5}$ is within the principal interval
$\Rightarrow\:\sin^{-1}\sin\large\frac{3\pi}{5}=\sin^{-1}\sin\large\frac{2\pi}{5}=\large\frac{2\pi}{5}$