${\sin }^{-1} \left(\sin \left(\frac{2x^2+4}{1+x^2}\right)\right)<\pi -3$ if

Let $t=\frac{2x^2+4}{1+x^2}\to x^2=\frac{4-t}{t-2}$

Since $x^2>0$ we get $2<t<4$

so we have to solve ${\sin }^{-1} (\sin t)<\pi -3$

and $2<t\le 4$ simultaneously

Now $\pi /2<2<t\le 4<3\pi /2$

So ${\sin }^{-1} (\sin t)=\pi -t$

and $\pi -t<\pi -3\Rightarrow t>3$

$\therefore$ we get $3<t\le 4$

$\Rightarrow 3<\frac{2x^2+4}{x^2+1}\le 4$

$\Rightarrow -1<x<1$ which is the required solution