$-1.3$
Notice,
$$
\begin{array}{l}
f(x)+f\left(\frac{1}{x}\right) \\
=\frac{1+2 x-x^{2}}{(1+x)\left(1+x^{2}\right)}+\frac{1+\frac{2}{x}-\frac{1}{x^{2}}}{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x^{2}}\right)} \\
=\frac{1+2 x-x^{2}}{(1+x)\left(1+x^{2}\right)}+\frac{x^{3}+2 x^{2}-x}{(1+x)\left(1+x^{2}\right)} \\
=\frac{x^{3}+x^{2}+x+1}{(1+x)\left(1+x^{2}\right)}=1
\end{array}
$$

Then the original expression
$$
\begin{array}{l}
= f(\tan \alpha)+f(\tan \beta)+f(\tan \gamma)+ \\
f\left(\frac{1}{\tan \alpha}\right)+f\left(\frac{1}{\tan \beta}\right)+f\left(\frac{1}{\tan \gamma}\right) \\
=3 .
\end{array}
$$
1. Given the function
$$
f(x)=\frac{1+2 x-x^{2}}{(1+x)\left(1+x^{2}\right)} \text {. }
$$

Let $\alpha, \beta, \gamma$ be the three interior angles of an arbitrary acute triangle. Then
$$
\begin{array}{l}
f(\tan \alpha)+f(\tan \beta)+f(\tan \gamma)+ \\
f(\cot \alpha)+f(\cot \beta)+f(\cot \gamma)=
\end{array}
$$