Given that, $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$
$$
\begin{aligned}
& s-a=11 k \\
& s-b=12 k \\
& s-c=13 k
\end{aligned}
$$

On adding,
$$
\begin{aligned}
3 s-(a+b+c) & =36 k \\
3 s-2 s & =36 k \\
s & =36 k
\end{aligned} \quad[\text { as } a+b+c=2 s]
$$

Now, $\tan ^2 \frac{A}{2}+\tan ^2 \frac{C}{2}$
$$
\begin{aligned}
& =\frac{(s-b)(s-c)}{s(s-a)}+\frac{(s-a)(s-b)}{s(s-c)} \\
& =\frac{(12 k)(13 k)}{(36 k)(11 k)}+\frac{(11 k)(12 k)}{(36 k)(13 k)}=\frac{12}{36}\left[\frac{13}{11}+\frac{11}{13}\right] \\
& =\frac{1}{3} \cdot\left[\frac{169+121}{11 \times 13}\right]=\frac{290}{429}
\end{aligned}
$$