Answer
$\color{blue}{56/65}$
Work Step by Step
Using a Sum Formula and substituting the given $\sin A = 3/5, \sin B = 5/13$ gives:
$$ \begin{align*}
\sin(A+B) &= \sin A\cos B + \sin B\cos A \\
\sin(A+B) &= (3/5)\cos B + (5/13)\cos A. \qquad (\text{Eq. 1})
\end{align*}$$
Since $\cos^2 B + \sin^2 B =1$ and $B\in QI$ (so that $\cos B \gt 0$),
$$\begin{align*} \cos B &= \sqrt{1 - \sin^2B} \\
&= \sqrt{1 - (5/13)^2} \\
&= \sqrt{1 - 25/169} \\
&= \sqrt{144/169} \\
\cos B &= 12/13.
\end{align*}$$
Similarly,
$$\begin{align*} \cos A &= \sqrt{1 - \sin^2A} \\
&= \sqrt{1 - (3/5)^2} \\
&= \sqrt{1 - 9/25} \\
&= \sqrt{16/25} \\
\cos A &= 4/5.
\end{align*}$$
Substituting the values obtained for $\cos A$ and $\cos B$ into (Eq. 1) above gives
$$\begin{align*}
\sin(A+B) &= (3/5)(12/13) + (5/13)(4/5) \\
&= 36/65 + 20/65 \\
\color{blue}{\sin(A+B)} &\color{blue}{= 56/65}
\end{align*}$$
