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
Given: $F=e^{-x} \sin y\hat{i}+e^{-y} \sin z\hat{j}+e^{-z} \sin x\hat{k}$
Definition of curl $F=\nabla \times F$
Let us consider $F=pi+qj+rk$
curl $F=(\dfrac{\partial r}{\partial x}-\dfrac{\partial q}{\partial z}){i}+(\dfrac{\partial p}{\partial z}-\dfrac{\partial r}{\partial x}){j}+(\dfrac{\partial q}{\partial x}-\dfrac{\partial p}{\partial y}){k}$
Thus, curl $F=-e^{-y} \cos z\hat{i}-e^{-z} \cos x\hat{j}-e^{-x} \cos y\hat{k}$
Definition of div $F=\nabla \cdot F$
Le us consider $F=pi+qj+rk$
That is, div $F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
Now, we have
div $F=\dfrac{\partial (e^{-x} \sin y)}{\partial x}+\dfrac{\partial (e^{-y} \sin z)}{\partial y}+\dfrac{\partial (e^{-z} \sin x)}{\partial z}$
or, div $F=-e^{-x} \sin y-e^{-y} \sin z-e^{-z} \sin x$
