Given $$0 < \alpha < {\pi \over 4}$$

and $$0 < \beta < {\pi \over 4}$$

$$ \therefore $$ $$0 > - \beta > - {\pi \over 4}$$

$$ \therefore $$ $$0 < \alpha + \beta < {\pi \over 2}$$

and $$ - {\pi \over 4} < \alpha - \beta < {\pi \over 4}$$

As cos($$\alpha $$ + $$\beta $$) = 3/5

so $${\tan \left( {\alpha + \beta } \right) = {4 \over 3}}$$

As sin( $$\alpha $$ - $$\beta $$) = 5/13

so $${\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}$$

Now tan(2$$\alpha $$) = tan($$\alpha $$ + $$\beta $$ + $$\alpha $$ - $$\beta $$)

= $${{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$$

= $${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$$ = $${{63} \over {16}}$$