Answer
$$\text{curl} \mathbf{(F+G)} = \operatorname{curl}(\mathbf{F})+\operatorname{curl}(\mathbf{G})$$
Work Step by Step
We expand and simplify:
\begin{align*}
\text{curl} \mathbf{(F+G)}&=\begin{array}{|||}
i&j &k\\
\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
\mathbf{F_1+G_1}& \mathbf{F_2+G_2}& \mathbf{F_3+G_3}
\end{array} \\
&= \left( \frac{\partial( \mathbf{F_3+G_3})}{\partial y}- \frac{\partial( \mathbf{F_2+G_2})}{\partial z}\right)i - \left( \frac{\partial( \mathbf{F_3+G_3})}{\partial x}- \frac{\partial(\mathbf{F_1+G_1})}{\partial z}\right)j + \left( \frac{\partial( \mathbf{F_2+G_2} )}{\partial x}- \frac{\partial( \mathbf{F_1+G_1} )}{\partial y}\right) k\\
&=\left[ \left( \frac{\partial( \mathbf{F_3})}{\partial y}- \frac{\partial( \mathbf{F_2})}{\partial z}\right)i - \left( \frac{\partial( \mathbf{F_3})}{\partial x}- \frac{\partial(\mathbf{F_1})}{\partial z}\right)j + \left( \frac{\partial( \mathbf{F_2} )}{\partial x}- \frac{\partial( \mathbf{F_1} )}{\partial y}\right) k\right]+
\left[ \left( \frac{\partial( \mathbf{G_3})}{\partial y}- \frac{\partial( \mathbf{G_2})}{\partial z}\right)i - \left( \frac{\partial( \mathbf{G_3})}{\partial x}- \frac{\partial(\mathbf{G_1})}{\partial z}\right)j + \left( \frac{\partial( \mathbf{G_2} )}{\partial x}- \frac{\partial( \mathbf{G_1} )}{\partial y}\right) k\right]\\
&= \operatorname{curl}(\mathbf{F})+\operatorname{curl}(\mathbf{G})
\end{align*}
