Explanation

(a)
We know that, $\sin{22\frac{1{^\circ}}{2} = \frac{1}{2}\sqrt{2 - \sqrt{2}}}$
and $\cos{22\frac{1{^\circ}}{2} = \frac{1}{2}\sqrt{2 + \sqrt{2}}}$
Since$,\ \alpha = 22{^\circ}30^{'} = 22\frac{1{^\circ}}{2}$
$\therefore\ \left( 1 + \cos{22\frac{1{^\circ}}{2}} \right)\left( 1 + \cos{67\frac{1{^\circ}}{2}} \right) \times \left( 1 + \cos{112\frac{1{^\circ}}{2}} \right)\left( 1 + \cos{157\frac{1{^\circ}}{2}} \right)$
$= \left( 1 + \frac{1}{2}\sqrt{2 + \sqrt{2}} \right)\left( 1 + \frac{1}{2}\sqrt{2 - \sqrt{2}} \right) \times \left( 1 - \frac{1}{2}\sqrt{2 - \sqrt{2}} \right)\left( 1 - \frac{1}{2}\sqrt{2 + \sqrt{2}} \right)$
$= \left\lbrack 1 - \frac{1}{4}(2 + \sqrt{2)} \right\rbrack\left\lbrack 1 - \frac{1}{4}(2 - \sqrt{2)} \right\rbrack$
$= \frac{\left( 2 - \sqrt{2} \right)(2 + \sqrt{2})}{16}$
$= \frac{4 - 2}{16} = \frac{1}{8}$