Explanation

(a)
Given, $\int_{\sin x}^{1}t^{2}\ f(t)dt = 1 - \sin x$
Now, $\frac{d}{dx}\int_{\sin x}^{1}t^{2}\ f(t)dt = \frac{d}{dx}(1 - \sin x)$
$\Rightarrow \left\lbrack 1^{2}f(1) \right\rbrack.(0) - (\sin^{2}{x)}.f\left( \sin x \right).\cos x = - \cos x$
[by Leibnitz formula]
$\Rightarrow$ Put $\sin x = 1/\sqrt{3\ }$
$\therefore\ f\left( \frac{1}{\sqrt{3}} \right) = \left( \sqrt{3} \right)^{2} = 3$