$\begin{array}{l}
\frac{a}{(x-b)}+\frac{b}{(x-a)}=2 \\
\Rightarrow\left[\frac{a}{(x-b)}-1\right]+\left[\frac{b}{(x-b)}-1\right]=0 \\
\Rightarrow \frac{a-(x-b)}{x-b}+\frac{b-(x-b)}{x-b}=0
\end{array}$
$\begin{array}{l}
\Rightarrow(a-x+b)\left[\frac{1}{(x-b)}+\frac{1}{(x-a)}\right]=0 \\
\Rightarrow(a-x+b)\left[\frac{(x-a)+(x-b)}{(x-b)(x-a)}\right]=0 \\
\Rightarrow(a-x+b)\left[\frac{2 x-(a+b)}{(x-b)(x-a)}\right]=0 \\
\Rightarrow(a-x+b)[2 x-(a+b)]=0 \\
\Rightarrow a-x+b=0 \text { or } 2 x-(a+b)=0 \\
\Rightarrow x=a+b \text { or } x=\frac{a+b}{2}
\end{array}$

Hence, the roots of the equation are $(a+b)$ and $\left(\frac{a+b}{2}\right)$.