(1)

$\sin x=\frac{-3}{5},$$\pi <x<\frac{3\pi }{2}$

$\cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-\frac{9}{25}}=\frac{-4}{5}$                 $[$$\because$ $x$ lies in third quadrant$]$

Now,  $\tan x=\frac{3}{4}$

So, $80({\tan }^{2}x-\cos x)=80(\frac{9}{16}+\frac{4}{5})=45+64=109$

(1)

$\frac{3\cos {36}^o+5\sin {18}^o}{5\cos {36}^o-3\sin {18}^o}$$=\frac{3\left[{\displaystyle \frac{1+\sqrt{5}}{4}}\right]+5\left[{\displaystyle \frac{\sqrt{5}-1}{4}}\right]}{5\left[{\displaystyle \frac{1+\sqrt{5}}{4}}\right]-3\left[{\displaystyle \frac{\sqrt{5}-1}{4}}\right]}$                                     $[\because \sin {18}^o=\frac{\sqrt{5}-1}{4} \mathrm{and} \cos {36}^o=\frac{1+\sqrt{5}}{4}]$

      $=\frac{3+3\sqrt{5}+5\sqrt{5}-5}{5+5\sqrt{5}-3\sqrt{5}+3}=\frac{8\sqrt{5}-2}{2\sqrt{5}+8}=\frac{4\sqrt{5}-1}{4+\sqrt{5}}$

      $=\frac{(4\sqrt{5}-1)(4-\sqrt{5})}{(\sqrt{5}+4)(4-\sqrt{5})}=\frac{16\sqrt{5}-20-4+\sqrt{5}}{16-5}$

      $=\frac{17\sqrt{5}-24}{11}=\frac{a\sqrt{5}-b}{c}$                              (Given)

$\therefore$   $a=17,$$b=24,$$c=11$

$\therefore$  $a+b+c=17+24+11=52$

(3)

As $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$=\frac{{\displaystyle \frac{1}{\sqrt{x(x^2+x+1)}}}+{\displaystyle \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}}{1-{\displaystyle \frac{1}{x^2+x+1}}}$

$=\frac{1+x}{\sqrt{x(x^2+x+1)}}\times \frac{x^2+x+1}{x^2+x}=\frac{1+x}{\sqrt{x}}\times \frac{\sqrt{x^2+x+1}}{x(1+x)}$

$=\sqrt{\frac{x^2+x+1}{x^3}}=\sqrt{x^{-1}+x^{-2}+x^{-3}}=\tan C$

$\Rightarrow A+B=C$