$\tan (2\alpha )=\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }=\frac{4}{1-4}=-\frac{4}{3}\Rightarrow \tan (3\alpha )=\tan (2\alpha +\alpha )=\frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }=\frac{-\frac{4}{3}+2}{1-\left( -\frac{4}{3}\times 2 \right)}=\frac{\frac{2}{3}}{\frac{11}{3}}=\frac{2}{11}$