Answer
$\color{blue}{33/65}$
Work Step by Step
Using a Sum Formula and substituting the given $\sin A = 3/5, \sin B = 5/13$ gives:
$$\begin{align*}
\cos(A+B) &= \cos A\cos B - \sin A\sin B \\
\cos(A+B) &= \color{green}{\cos A}\color{magenta}{\cos B} - (3/5)(5/13). \qquad (\text{Eq. 1})
\end{align*}$$
We need $\color{magenta}{\cos A}$ and $\color{green}{\cos B}$ to complete the answer.
Since $\cos^2 A + \sin^2 A =1$ and $A\in QI$ (so that $\cos A \gt 0$),
$$\begin{align*}
\color{green}{\cos A}\ &= \sqrt{1 - \sin^2A} \\
&= \sqrt{1 - (3/5)^2} \\
&= \sqrt{1 - 9/25} \\
&= \sqrt{16/25} \\
\color{green}{\cos A}\ &\color{green}{= 4/5}
\end{align*}$$
Similarly, for $\cos B,\ B \in QI$,
$$\begin{align*}
\color{magenta}{\cos B}\ &= \sqrt{1 - \sin^2B} \\
&= \sqrt{1 - (5/13)^2} \\
&= \sqrt{1 - 25/169} \\
&= \sqrt{144/169} \\
\color{magenta}{\cos B}\ &\color{magenta}{= 12/13}
\end{align*}$$
Substituting the values obtained for $\cos A$ and $\cos B$ into (Eq. 1) above gives
$$\begin{align*}
\cos(A+B) &= \color{green}{(4/5)}\color{magenta}{(12/13)} - (3/5)(5/13) \\
&= 48/65 - 15/65 \\
\color{blue}{\cos(A+B)} &\color{blue}{= 33/65}
\end{align*}$$