Explanation

(a)
Here, $\sin{18{^\circ} + \cos^{2}{36{^\circ}}}$
$= \left( \frac{\sqrt{5} - 1}{4} \right)^{4} + \left( \frac{\sqrt{5} + 1}{4} \right)^{2}$
$= \frac{5 + 1 - 2\sqrt{5}}{16} + \frac{5 + 1 + 2\sqrt{5}}{16}$
$= \frac{12}{16} = \frac{3}{4}$
And $\sin^{2}{18{^\circ}.\cos^{2}{36{^\circ} = \left( \frac{\sqrt{5} - 1}{4} \right)^{2}\left( \frac{\sqrt{5 + 1}}{4} \right)^{2}}}$
$= \left( \frac{5 - 1}{4 \times 4} \right)^{2} = \frac{1}{16}$
Required equation is
$x^{2} -$(sum of roots)$x +$(products of roots)=0
$\Rightarrow \ \ \ x^{2} - \frac{3}{4}x + \frac{1}{16} = 0$
$\Rightarrow \ \ 16x^{2} - 12x + 1 = 0$