Official solution
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(\[\alpha \] + \[\beta \] + \[\alpha \] - \[\beta \])
= \[{{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}\]
= \[{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}\] = \[{{63} \over {16}}\]