2. Let $f(x)=\frac{a^{x}}{1+a^{x}}(a>0, a \neq 1), [m]$ denote the greatest integer not exceeding the real number $m$, then the range of the function $\left[f(x) -\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]$ is $\qquad$ .
2. $\{-1,0\}$.

Let $g(x)=f(x)-\frac{1}{2}$, then
$$
\begin{aligned}
g(-x) & =\frac{a^{-x}}{1+a^{-x}}-\frac{1}{2}=\frac{1}{a^{x}+1}-\frac{1}{2} \\
& =\frac{\left(1+a^{x}\right)-a^{x}}{1+a^{x}}-\frac{1}{2} \\
& =\frac{1}{2}-\frac{a^{x}}{1+a^{x}} \\
& =-g(x) .
\end{aligned}
$$
$$
\begin{array}{ll}
\therefore \quad & {\left[f(x)-\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]} \\
& =[g(x)]+[-g(x)] . \\
\because \quad & a^{x}>0, \\
\therefore \quad & 0<f(x)=\frac{a^{x}}{1+a^{x}}<1 . \\
\therefore \quad & -\frac{1}{2}<g(x)=f(x)-\frac{1}{2}<\frac{1}{2} .
\end{array}
$$

When $-\frac{1}{2}<g(x)<0$,
$$
[g(x)][-g(x)]=-1+0=-1 \text {; }
$$

When $g(x)=0$, $[g(x)]+[-g(x)]=0$;
When $0<g(x)<\frac{1}{2}$,
$$
[g(x)]+\{-g(x)]=0+(-1)=-1 \text {. }
$$

In summary, the range of the function $\left[f(x)-\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]$ is $\{-1,0\}$.