$\begin{aligned} & \because Q(x * 1) * 1=x *(1 * 1) \\ & \left(x^2+1\right) * 1=x *(2) \\ & \left(x^2+1\right)^2+1=x^2+8 \\ & x^4+x^2-6=0 \Rightarrow\left(x^2+3\right)\left(x^2-2\right)=0 \\ & x^2=2 \\ & \Rightarrow 2 \sin ^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2 \sin ^{-1}\left(\frac{1}{2}\right) \\ & =\frac{\pi}{3}\end{aligned}$