Explanation

(a)
Here, $f(x) = \sqrt{\sin^{- 1}{(2x) + \frac{\pi}{6}}}$ , to find domain we must have,
$\sin^{- 1}{(2x) + \frac{\pi}{6} \geq 0}$
$\ \ \ \ \left( but - \frac{\pi}{2} \leq \sin^{- 1}\theta \leq \frac{\pi}{2} \right)$
$\therefore\ \ \ \ \ \ \ \ \ \ - \frac{\pi}{6} \leq \sin^{- 1}{(2x)} \leq \frac{\pi}{2}$
${\Rightarrow sin}\left( - \frac{\pi}{6} \right) \leq 2x \leq \sin\left( \frac{\pi}{2} \right)$
$\Rightarrow \ - \frac{1}{2} \leq 2x \leq 1\ \Rightarrow x \in \left\lbrack - \frac{1}{4},\ \frac{1}{2} \right\rbrack$