Explanation

(a)
Since, $\alpha = \sin^{- 1}{\left( \frac{4}{5} \right) + \sin^{- 1}\left( \frac{1}{3} \right)}$
$= \sin^{- 1}\left( \frac{4}{5}\sqrt{1 - \frac{1}{9}} + \frac{1}{3}\sqrt{1 - \frac{16}{25}} \right)$
$\Rightarrow \alpha = \sin^{- 1}{\left( \frac{8\sqrt{2}}{15} + \frac{3}{15} \right) = \sin^{- 1}\ \left( \frac{8\sqrt{2} + 3}{15} \right)}$
Since, $\frac{8\sqrt{2} + 3}{15} < 1$
$\therefore\alpha < \frac{\pi}{2}$
Now, $\beta = \cos^{- 1}{\left( \frac{4}{5} \right) + \cos^{- 1}\left( \frac{1}{3} \right)}$
$\Rightarrow \beta = \frac{\pi}{2} - \sin^{- 1}{\left( \frac{4}{5} \right) + \frac{\pi}{2} - \sin^{- 1}\left( \frac{1}{3} \right)}$
$= \pi - \left( \sin^{- 1}{\frac{4}{5} + \sin^{- 1}\frac{1}{3}} \right)$
$= \pi - \alpha$
$\Rightarrow \beta > \alpha$ $\left( \therefore\alpha < \frac{\pi}{2} \right)$