Answer
curl $F=-e^{-y} \cos z\hat{i}-e^{-z} \cos x\hat{j}-e^{-x} \cos y\hat{k}$
and
div $F=-e^{-x} \sin y-e^{-y} \sin z-e^{-z} \sin x$
Work Step by Step
Use definition of curl $F=\nabla \times F$
Consider $F=ai+bj+ck$
That is, curl $F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$
Here, we have $F=e^{-x} \sin y\hat{i}+e^{-y} \sin z\hat{j}+e^{-z} \sin x\hat{k}$
Thus, curl $F=-e^{-y} \cos z\hat{i}-e^{-z} \cos x\hat{j}-e^{-x} \cos y\hat{k}$
Use definition of div $F=\nabla \times F$
Consider $F=ai+bj+ck$
That is, div $F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
Here, we have $F=e^{-x} \sin y\hat{i}+e^{-y} \sin z\hat{j}+e^{-z} \sin x\hat{k}$
Thus, div $F=\dfrac{\partial e^{-x} \sin y}{\partial x}+\dfrac{\partial e^{-y} \sin z}{\partial y}+\dfrac{\partial e^{-z} \sin x}{\partial z}=-e^{-x} \sin y-e^{-y} \sin z-e^{-z} \sin x$
