Let $S=50\tan \left(3{\tan }^{-1}\frac{1}{2}+2{\cos }^{-1}\frac{1}{\sqrt{5}}\right)$$+4\sqrt{2}\tan \left(\frac{1}{2}{\tan }^{-1}2\sqrt{2}\right)$

$=50\tan \left({\tan }^{-1}\frac{1}{2}+2\left({\tan }^{-1}\frac{1}{2}+{\tan }^{-1}2\right)\right)$$+4\sqrt{2}\tan \left(\frac{1}{2}{\tan }^{-1}2\sqrt{2}\right)$

Let $\frac{1}{2}{\tan }^{-1}2\sqrt{2}=\alpha \Rightarrow \tan 2\alpha =2\sqrt{2}$

Now $\frac{2\tan \alpha }{1-{\tan }^2\alpha }=2\sqrt{2}$

Solving we get, $\tan \alpha =\frac{1}{\sqrt{2}}$

So$\left.S=50\tan \left({\tan }^{-1}\frac{1}{2}+2·\frac{\pi }{2}\right)\right)+4\sqrt{2}\times \frac{1}{\sqrt{2}}$

$=50\left({\mathrm{tantan}}^{-1}\frac{1}{2}\right)+4$

$=25+4=29$