1. Let $f(x)=x^{10}+2 x^{9}-2 x^{8}-2 x^{7}+x^{6}$ $+3 x^{2}+6 x+1$, then $f(\sqrt{2}-1)=$
$=1.4$.
Let $x=\sqrt{2}-1$, then $x+1=\sqrt{2} \Rightarrow (x+1)^{2}=2 \Rightarrow x^{2}+2x-1=0$. That is, $x=\sqrt{2}-1$ is a root of $x^{2}+2x-1=0$. But $f(x)=(x^{8}-x^{6}+3)(x^{2}+2x-1)+4$, so $f(\sqrt{2}-1)=4$.