1. Given $f(x)=\operatorname{tg}\left(\frac{\pi}{4}-\operatorname{arctg} x\right)$. Then $f\left(\frac{1-\sqrt[3]{3}-\sqrt{2}}{1+\sqrt[3]{3}+\sqrt{2}}\right)=$ $\qquad$ 2.
$$
=\therefore \sqrt[3]{3}+\sqrt{2} \text {. }
$$

Given the function simplified as $f(x)=\frac{1-x}{1+x}$. By the properties of ratios, we get $\quad \frac{1-f(x)}{r f(x)} x$, that is $\quad f[f(x)]=\ldots$ Thus, $f\left(\frac{1-\sqrt[3]{3} \cdot \sqrt{2}}{1+\sqrt[3]{3}+\sqrt{2}}\right.$ $\Rightarrow f[f(\sqrt[3]{3}+\sqrt{2})]=\sqrt[3]{3}+\sqrt{2}$.