Explanation

(a)
Roots of equation $x^{2} - 9x + 8 = 0\ are\ 1\ and\ 8$
Let $y = \lbrack\sin^{2}{\alpha + \sin^{4}{\alpha + \sin^{6}{\alpha + ...\infty\rbrack\log_{e}2}}}$
$\Rightarrow y = \frac{\sin^{2}\alpha}{1 - \sin^{2}\alpha}\log_{e}{2 = \tan^{2}{\alpha\log_{e}2}}$
$\Rightarrow y = \log_{e}2^{\tan^{2}\alpha}$
$\Rightarrow e^{y} = 2^{\tan^{2}\alpha}$
According to question,
$2^{\tan^{2}\alpha} = 8 = 2^{3} \Rightarrow \tan^{2}{\alpha = 3}$
$\Rightarrow \tan{\alpha = \sqrt{3} \Rightarrow \alpha = \frac{\pi}{3}}$
$\therefore\sin^{- 1}\left( \sin\frac{2\pi}{3} \right) = \pi - \frac{2\pi}{3} = \frac{\pi}{3} = \alpha$