$1=\cos 3A+\cos 3B+\cos 3C=2\cos \frac{3(A+B)}{2}\cos \frac{3(A-B)}{2}+1-2\sin^2 \frac{3C}{2}$
$\Rightarrow 1=-2\sin \frac{3C}{2}\cos \frac{3(A-B)}{2}=1-2\sin \frac{3C}{2}\left(\cos \frac{3(A+B)}{2}-\cos \frac{3(A-B)}{2} \right)$
$\Rightarrow 1=1+4\sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}\Rightarrow \sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0$
$\Rightarrow \left[\begin{array}{l}\angle A=120^o\\\angle B=120^o\\\angle C=120^o\end{array}\right. \Rightarrow \angle A=120^o$, vì $\angle A>\angle B>\angle C$
$0=\sin 5A+\sin 5B+\sin 5C=2\sin \frac{5(A+B)}{2}\cos \frac{5(A-B)}{2}+2\sin \frac{5C}{2}\cos \frac{5C}{2}$
$=2\cos \frac{5C}{2}\left(\cos \frac{5(A-B)}{2}+\cos \frac{5C}{2} \right)$
$\Rightarrow \cos \frac{5A}{2}\cos \frac{5B}{2}\cos \frac{5C}{2}=0$
$\Rightarrow \left[\begin{array}{l}\angle A=36^o\\\angle B=36^o\\\angle C=36^o\end{array}\right. \Rightarrow \angle B=36^o$, vì $\angle A>\angle B>\angle C$
Từ đó suy ra $\angle A=120^o,\angle B=36^o,\angle C=24^o$