<s>[INST] The sum of angles in a triangle is greater than 180? So basically the question goes:
Let $ABC$ be a triangle and that $\cos A=-\frac 3 5$ and $\cos B=\frac 5 {13}$, find $\cos C$. 
What I did was I get the value of $A$ and $B$ by calculating the arccosine of $-3/5$ and $5/13$ and the sum of both of these angles was already greater than $180$, how can I calculate $\cos C$?
 [/INST] I suspect you're meant to calculate $\cos C$ by using the cosine of sums rule, not by computing inverse cosines.

With that in mind, and on the off chance that the minus sign on $-3/5$ is an error, and the cosine of angle $A$ is really meant to be $3/5$, I'll point out
$$
\arccos 3/5 \approx 53.13 \\
\arccos 5/13 \approx 67.38
$$
which leaves $C \approx 180-53.13-67.38 = 59.49$.  And indeed, we see that (assuming the minus sign is spurious)
\begin{align}
\cos(A+B) & = \cos A \cos B - \sin A \sin B \\
          & = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} \\
          & = \frac{15}{65} - \frac{48}{65} = -\frac{33}{65}
\end{align}
That would mean that $\cos C = 33/65$ (because $A+B+C = 180$), and sure enough,
$$
\arccos 33/65 \approx 59.49
$$
</s>