Given that $\sin \alpha -\cos \alpha =m$ and $\sin 2\alpha =n-m^2$, where $-\sqrt{2}\le m\le \sqrt{2}$, 

We know that, $\sin 2\alpha =2\sin \alpha \cos \alpha$

$\Rightarrow 2\sin \alpha \cos \alpha =n-{\left(\sin \alpha -\cos \alpha \right)}^2$

$\Rightarrow 2\sin \alpha \cos \alpha =n-{\sin }^2\alpha -{\cos }^2\alpha +2\sin \alpha \cos \alpha$

$\Rightarrow n={\sin }^2\alpha +{\cos }^2\alpha$

Now, We know that, ${\sin }^2\alpha +{\cos }^2\alpha =1$

$\Rightarrow n=1$