2. If real numbers $a, b (a \neq b)$ satisfy $f(x)=$ $-\frac{x+a}{x+b}$, and the inverse function $F(x)$ has a center of symmetry $M$, then the coordinates of point $M$ are
2. $(-1,-b)$.
First, $F(x)=\frac{-a-b x}{1+x}=-b+\frac{b-a}{1+x}$.
Let $M\left(x_{0}, y_{0}\right)$, then for any $x_{1}, x_{2} \in \mathbf{R}$, if $x_{1}+x_{2}=2 x_{0}$, we have $F\left(x_{1}\right)+F\left(x_{2}\right)=2 y_{0}$, that is
$$
\begin{array}{l}
(b-a)\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}\right)=2\left(y_{0}+b\right) \\
=(b-a) \frac{2\left(1+x_{0}\right)}{\left(1+x_{1}\right)\left(1+x_{2}\right)}
\end{array}
$$

Therefore, $1+x_{0}=0, y_{0}+b=0$, which gives $M(-1,-b)$.