Given:

$\sin B=\frac{5}{13}\Rightarrow \tan B=\frac{5}{12}$

Now,

$\sin \left(A-B\right)=\frac{16}{65}$

$\Rightarrow \tan \left(A-B\right)=\frac{16}{63}$

$\Rightarrow \frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{16}{63}$

$\Rightarrow \frac{\tan A-{\displaystyle \frac{5}{12}}}{1+{\displaystyle \frac{5\tan A}{12}}}=\frac{16}{63}$

$\Rightarrow \frac{12\tan A-{\displaystyle 5}}{12+{\displaystyle 5\tan A}}=\frac{16}{63}$

$\Rightarrow 756\tan A-315=192+80\tan A$

$\Rightarrow 676\tan A=507$

$\Rightarrow \tan A=\frac{507}{676}$

So,

$\tan A+\cot A=\frac{507}{676}+\frac{676}{507}=\frac{714025}{342732}$