Answer
$curl (F+G)=curl F+curl G$
Work Step by Step
The vector field $F$ is conservative when $curl F=0$
When $F=ai+bj+ck$, then we have $curl F=[c_y-b_z]i+[a_z-c_z]j+[b_x-a_y]k$
Plug $F=a_1i+b_1j+c_1z; G=a_2i+b_2j+c_2k$
and $curl (F+G)=curl [(a_1+a_2)i+(b_1+b_2)j+(b_3+c_3)k$
Apply distributive property of the cross product in the $curl (F+G)=curl [(a_1+a_2)i+(b_1+b_2)j+(b_3+c_3)k$.
This implies that $curl (F+G)=\nabla \times F+\nabla \times G=curl F+curl G$
Hence, $curl (F+G)=curl F+curl G$ (proved)
