The domain of  ${\sin }^{-1}\left[{\log }_2(x^2/2)\right]$ is 

$$\begin{gathered} We know that domain of {\sin }^{-1}x is \left[-1,1\right]. \\ \therefore {\sin }^{-1}\left[{\log }_2\left(\frac{x^2}{2}\right)\right]is defined if-1\le {\log }_2\left(\frac{x^2}{2}\right)\le 1 \\ \Rightarrow 2^{-1}\le \frac{x^2}{2}\le 2 \\ \Rightarrow 1\le x^2\le 4 \\ \Rightarrow x\in \left[-2,-1\right]\cup \left[1,2\right] \end{gathered}$$