Given,

Pair of straight lines equation $a{x}^2-4xy-2y^2=0$

Now angle between them is given by $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

Now putting the value in formula we get,

$\tan \theta =\frac{2\sqrt{4+2a}}{a-2}$

Now given $\cos \theta =\frac{1}{5}\Rightarrow \tan \theta =\frac{\sqrt{25-1}}{1}=\sqrt{24}$

$\Rightarrow \frac{2\sqrt{4+2a}}{a-2}=\sqrt{24}$

$\Rightarrow 2\sqrt{4+2a}=\sqrt{24}\left(a-2\right)$

$\Rightarrow 4\left(4+2a\right)=24{\left(a-2\right)}^2$

$\Rightarrow \left(4+2a\right)=6{\left(a-2\right)}^2$

$\Rightarrow 6a^2-24a+24-4-2a=0$

$\Rightarrow 6a^2-26a+20=0$

$\Rightarrow 6a^2-6a-20a+20=0$

$\Rightarrow 6a\left(a-1\right)-20\left(a-1\right)=0$

$\Rightarrow \left(6a-20\right)\left(a-1\right)=0$

$\Rightarrow a_1=1 \& a_2=\frac{20}{6}=\frac{10}{3}$

So, $a_1+3a_2=1+10=11$