Finding $\cos(A+C)$ given that $A,B,C$ are angles in an acute triangle. Given that $A,B,C$ be angles in an acute triangle.
If $(5+4\cos A)(5-4\cos B)=9$ and $(13-12\cos B)(13-12\cos C)=25$
find $\cos(A+C)$. 
I know $A+B+C=180^\circ$ and $\cos B=-\cos(A+C)$ and what next?

Thanks for the challenging problem! I enjoyed working through it.
The tangent half-angle substitution works really well here. Let $a = \tan ({A\over 2})$, $b = \tan({B\over2})$, $c = \tan({C\over2})$ then $a,b,c > 0$ and
$$ \cos A = \frac{1-a^2}{1+a^2}, \quad \cos B = \frac{1-b^2}{1+b^2}, \quad \cos C = \frac{1-c^2}{1+c^2} $$
Substituting these into the given equations and simplifying we (eventually) get $$ a = 3b, \quad b = {1\over 5c}$$
Using the triangle relationship we also have 
$$ \tan \left( {A\over2} + {B\over2} \right) = \tan \left(90^\circ-{C\over2}\right) = \cot \left({C\over2}\right) $$
$$ \frac{a+b}{1-ab} = \frac{1}{c} $$
Thus
$$  \frac{4b}{1-3b^2} = 5b $$
$$ b^2 = {1\over 15}$$
Back-substituting gives
$$ \cos B = -\cos(A+C) = {7\over 8} $$
