Answer
a) $(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k$
b) $0$
Work Step by Step
a) Consider $F=A i+B j+C k$
Then $curl F=\begin{vmatrix}i&j&k\\\dfrac{\partial}{\partial x}&\dfrac{\partial }{\partial y}&\dfrac{\partial }{\partial z}\\A&B&C\end{vmatrix}$
$curl F=[C_y-B_z]i+[A_z-C_z]j+[B_x-A_y]k$
$curl F=(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k=(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k$
b) $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$
$div F=\dfrac{\partial (\sin yz)}{\partial x}+\dfrac{\partial (\sin zx)}{\partial y}+\dfrac{\partial (\sin xy)}{\partial z}=0$
