Statement –I: The equation ${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3-a{\pi }^3=0$ has a solution for all  $a\ge \frac{1}{32}$.
Statement –II: For any  $x\in [-1,1],{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2},$ and  $0\le {({\sin }^{-1}x-\frac{\pi }{4})}^2\le \frac{9{\pi }^2}{16}$

$\begin{array}{l}{\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]\\ \Rightarrow -\frac{3\pi }{4}\le ({\sin }^{-1}x-\frac{\pi }{4})\le \frac{\pi }{4}\\ 0\le {({\sin }^{-1}x-\frac{\pi }{4})}^2\le \frac{9}{16}{\pi }^2\mathrm{........}\left(i\right)\\ StatementIIistrue\\ {\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3=a{\pi }^3\\ \Rightarrow ({\sin }^{-1}x+{\cos }^{-1}x)\left[{({\sin }^{-1}x+{\cos }^{-1}x)}^2-3{\sin }^{-1}x{\cos }^{-1}x\right]=a{\pi }^3\\ \Rightarrow \frac{{\pi }^2}{4}-3{\sin }^{-1}x{\cos }^{-1}x=2a{\pi }^2\\ \Rightarrow {({\sin }^{-1}x-\frac{\pi }{4})}^2=\frac{{\pi }^2}{12}(8a-1)+\frac{{\pi }^2}{16}\\ \Rightarrow {({\sin }^{-1}x-\frac{\pi }{4})}^2=\frac{{\pi }^2}{48}(32a-1)\\ Puttingthisvalueinequation\left(i\right)\\ \Rightarrow 0\le 32a-1\le 27\\ \frac{1}{32}\le a\le \frac{7}{8}\\ Statement1isfalse\end{array}$