Answer
$curl (F+G)=curl F+curl G$
Work Step by Step
Need to prove that $div (F+G)=div F+div G$
The vector field $F$ will be conservative if and only if $curl F=0$
consider $F=A i+B j+C k$
$curl F=[C_y-B_z]i+[A_z-C_z]j+[B_x-A_y]k$
Suppose $F=a_1i+b_1j+c_1z; G=a_2i+b_2j+c_2k$
and $curl (F+G)=curl [(a_1+a_2)i+(b_1+b_2)j+(b_3+c_3)k$
By using the distributive property of the cross product, we have
$curl (F+G)=\nabla \times F+\nabla \times G=curl F+curl G$
Hence, $curl (F+G)=curl F+curl G$
