If tan A

$\tan A=\frac{1}{2}, \quad \tan B=\frac{1}{3}$

then $\tan (2 A+B)=\frac{\tan 2 A+\tan B}{1-\tan 2 A \tan B}$

Using Identities :

$\left(\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}\right)$

$=\frac{\left(\frac{2 \tan A}{1-\tan ^{2} A}\right)+\tan B}{1-\left(\frac{2 \tan A}{1-\tan ^{2} A}\right) \tan B} \quad$ (using identity $\left.\tan 2 a=\frac{2 \tan a}{1-\tan ^{2} a}\right)$

$=\frac{\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}}+\frac{1}{3}}{1-\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}} \times \frac{1}{3}}$

$=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3} \times \frac{1}{3}}$

$=\frac{5 / 3}{1-4 / 9}$

$=\frac{5 / 3}{\frac{9-4}{9}}=\frac{5}{3} \times \frac{9}{5}$

$\therefore \tan (2 A+B)=3$

Hence, the correct answer is option C.