$\sin^{-1} (\sin \frac{3\pi}{5}) = \dots \dots \dots$

$\tan^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \sin^{-1} \frac{1}{\sqrt{2}} \right) \right] = \dots$

Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $a=3, b=4$ and $A=\sin^{-1}\left(\frac{3}{4}\right)$,then the angle $B$ is (in $^{\circ}$)

If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$,then $x=$

The real part of $\sin^{-1}(e^{i\theta})$ is

Find the value of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$.