Answer
$$\operatorname{div}(\mathbf{F}+\mathbf{G})=\operatorname{div}(\mathbf{F})+\operatorname{div}(\mathbf{G})$$
Work Step by Step
We expand and simplify:
\begin{align*}
\operatorname{div}(\mathbf{F}+\mathbf{G})&= \frac{\partial( \mathbf{F}_x+\mathbf{G}_x)}{\partial x}+\frac{\partial( \mathbf{F}_y+\mathbf{G}_y) }{\partial y}+\frac{\partial ( \mathbf{F}_z+\mathbf{G}_z)}{\partial z}\\
&= \frac{\partial( \mathbf{F}_x )}{\partial x}+ \frac{\partial( \mathbf{G}_x)}{\partial x}+\frac{\partial( \mathbf{F}_y ) }{\partial y}+\frac{\partial( \mathbf{G}_y) }{\partial y}+\frac{\partial ( \mathbf{F}_z )}{\partial z}+\frac{\partial ( \mathbf{G}_z)}{\partial z}\\
&=\left( \frac{\partial( \mathbf{F}_x )}{\partial x}+\frac{\partial( \mathbf{F}_y ) }{\partial y}+\frac{\partial ( \mathbf{F}_z )}{\partial z}\right)+\left( \frac{\partial( \mathbf{G}_x)}{\partial x}+\frac{\partial( \mathbf{G}_y) }{\partial y}+\frac{\partial ( \mathbf{G}_z)}{\partial z} \right) \\
&=\operatorname{div}(\mathbf{F})+\operatorname{div}(\mathbf{G})
\end{align*}
