已知 $\sqrt{2}<a<\pi$, 化简: $|a-\pi|+|a-\sqrt{2}|-\sqrt{(2-\pi)^{2}}=$
$\because \sqrt{2}<a<\pi, \quad \therefore|a-\pi|+|a-\sqrt{2}|-\sqrt{(2-\pi)^{2}}=\pi-a+a-\sqrt{2}-(\pi-2)=2-\sqrt{2}$.

故答案为: $2-\sqrt{2}$.