Answer
a) $ze^xi+(xye^z-yze^x)j-xe^zk$
b) $y(e^z+e^x)$
Work Step by Step
a) When $F=ai+bj+ck$, then we have $curl F=[c_y-b_z]i+[a_z-c_z]j+[b_x-a_y]k$
$curl F=[ze^x-0]i+[xye^z-yze^x]j+[0-xe^z]k=ze^xi+(xye^z-yze^x)j-xe^zk$
b) $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}=\dfrac{\partial (xye^z)}{\partial x}+\dfrac{\partial (0)}{\partial y}+\dfrac{\partial (yze^x)}{\partial z}=y(e^z+e^x)$
