$\begin{array}{l}
\frac{x-4}{x-5}+\frac{x-6}{x-7}=3 \frac{1}{3}, x \neq 5,7 \\
\Rightarrow \frac{(x-4)(x-7)+(x-5)(x-6)}{(x-5)(x-7)}=\frac{10}{3} \\
\Rightarrow \frac{x^{2}-11 x+28+x^{2}-11 x+30}{x^{2}-12 x+35}=\frac{10}{3} \\
\Rightarrow \frac{2 x^{2}-22 x+58}{x^{2}-12 x+35}=\frac{10}{3}
\end{array}$
$\begin{array}{l}
\Rightarrow \frac{x^{2}-11 x+29}{x^{2}-12 x+35}=\frac{5}{3} \\
\Rightarrow 3 x^{2}-33 x+87=5 x^{2}-60 x+175 \\
\Rightarrow 2 x^{2}-27 x+88=0 \\
\Rightarrow 2 x^{2}-16 x-11 x+88=0 \\
\Rightarrow 2 x(x-8)-11(x-8)=0 \\
\Rightarrow(x-8)(2 x-11)=0 \\
\Rightarrow x-8=0 \text { or } 2 x-11=0 \\
\Rightarrow x=8 \text { or } x=\frac{11}{2}
\end{array}$

Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.