Let A, B, C be angles of a triangle with ${\cos }^{2}A+{\cos }^{2}B+2\sin A\sin B\cos C=\frac{15}{8}$  and  ${\cos }^{2}B+{\cos }^{2}C+2\sin B\sin C\cos A=\frac{14}{9}$ 
Which is/are CORRECT?

In triangle ABC  ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C+2\cos A\cos B\cos C=1$
 ${\cos }^{2}A+{\cos }^{2}B+2\sin A\sin B\cos C=\frac{15}{8}$  ________(1)
${\cos }^{2}B+{\cos }^{2}C+2\sin B\sin C\cos A=\frac{14}{9}$  _________(2)
Add  ${\cos }^{2}C+2(\cos A\cos B-\sin A\sin B)\cos C$  to both sides of the first given equation.
 $1=\frac{15}{8}-2{\cos }^{2}C+{\cos }^{2}C$
so cosC is  $\sqrt{\frac{7}{8}}$ and therefore sin C is $\sqrt{\frac{1}{8}}$
Similarly, we have $\cos A = \sqrt{\frac{14}{9}-1}=\sqrt{\frac{5}{9}}$ 

${\cos }^{2}C+{\cos }^{2}A+2\sin C\sin A\cos B=x$   ______(3)
Add 1 & 3, we get
${\cos }^{2}B+{\cos }^{2}C=x-\frac{1}{8}$           _______(4)
Add 2 & 3, we get
${\cos }^{2}A+{\cos }^{2}B=x-\frac{4}{9}$___________(5)
$\sin B=\sin (A+C)$  can be computed first and then  ${\cos }^{2}B$  is easily found.
 ${\cos }^{2}C+{\cos }^{2}A+2\sin C\sin A\cos B=\frac{111-4\sqrt{35}}{72}$