Explanation

(a)
Given, $e^{\sin x} - e^{- \sin x} - 4 = 0$
$\Rightarrow \ e^{2\sin x} - 4e^{\sin x} - 1 = 0$
$\Rightarrow \ e^{\sin x} = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 + \sqrt{5}$
$\Rightarrow \ \sin x = \log{(2 + \sqrt{5})}\ \ \ \lbrack\because\ \log\left( 2 - \sqrt{5} \right)\ is\ not\ defined\rbrack$
Since, $2 + \sqrt{5} > e\ \Rightarrow \ \left( 2 + \sqrt{5} \right) > 1$
$\Rightarrow \sin{x > 1}$, which is not possible
Hence, no solution exist