$\tan \left( \frac{\pi }{4}-\alpha  \right)=\frac{1}{5}\Rightarrow \frac{1-\tan \alpha }{1+\tan \alpha }=\frac{1}{5}$ $\Rightarrow 5-5\tan \alpha =1+\tan \alpha \Rightarrow 4=6\tan \alpha \Rightarrow \tan \alpha =\frac{2}{3}$ $\tan 2\alpha =\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\Rightarrow \tan 2\alpha =\frac{2\left( \frac{2}{3} \right)}{1-\frac{4}{9}}=\frac{12}{5}=2/4$