Correct option: A.Both statements $I$ and $II$ are true.
a
${\sin ^{ - 1}}x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$
$ \Rightarrow - \frac{{3\pi }}{4} \le \left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right) \le \frac{\pi }{4}$
$0 \le {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} \le \frac{9}{{16}}{\pi ^2}\,\,\,\,\,\,\,.....\left( 1 \right)$
Statement $II$ is true
${\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3} = a{\pi ^3}$
$\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)\left[ {{{\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)}^2} - 3{{\sin }^{ - 1}}x{{\cos }^{ - 1}}x} \right] = a{\pi ^3}$
$ \Rightarrow \frac{{{\pi ^2}}}{4} - 3{\sin ^{ - 1}}x{\cos ^{ - 1}}x = 2a{\pi ^2}$
$ \Rightarrow {\sin ^{ - 1}}x\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right) = \frac{{{\pi ^2}}}{{12}}\left( {1 - 8a} \right)$
$ \Rightarrow {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} = \frac{{{\pi ^2}}}{{12}}\left( {1 - 8a} \right) + \frac{{{\pi ^2}}}{{16}}$
$ \Rightarrow {\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} = \frac{{{\pi ^2}}}{{48}}\left( {32a - 1} \right)$
Putting this value in equation $(1)$
$0 \le \frac{{{\pi ^2}}}{{48}}\left( {32a - 1} \right) \le \frac{9}{{16}}{\pi ^2}$
$ \Rightarrow 0 \le 32a - 1 \le 27$
$\frac{1}{{32}} \le a \le \frac{7}{8}$
Statement - $I$ is also true.