Answer
a)$\vec{\nabla{(fg)}}=f\vec{\nabla g}+g\vec{\nabla f}$
b)$\vec{\nabla}\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\vec{\nabla}\times\vec{A})-\vec{A}\cdot(\vec{\nabla}\times\vec{B})$
c)$\vec{\nabla}\times(f\vec{A})=f(\vec{\nabla}\times\vec{A})-\vec{A}\times(\vec{\nabla f})$
Work Step by Step
a)$\vec{\nabla{(fg)}}=\hat{x}\frac{\partial}{\partial x}(fg)+\hat{y}\frac{\partial}{\partial y}(fg)+\hat{z}\frac{\partial}{\partial z}(fg)$
Using chain rule gives,
$\frac{\partial}{\partial x}(fg)=f\frac{\partial}{\partial x}(g)+g\frac{\partial}{\partial x}(f)$
Similarly,
$\frac{\partial}{\partial y}(fg)=f\frac{\partial}{\partial x}(g)+g\frac{\partial}{\partial y}(f)$
$\frac{\partial}{\partial z}(fg)=f\frac{\partial}{\partial z}(g)+g\frac{\partial}{\partial z}(f)$
$\therefore \vec{\nabla{(fg)}}=\hat{x}(f\frac{\partial}{\partial x}(g)+g\frac{\partial}{\partial x}(f))+\hat{y}(f\frac{\partial}{\partial y}(g)+g\frac{\partial}{\partial y}(f))+\hat{z}(f\frac{\partial}{\partial z}(g)+g\frac{\partial}{\partial z}(f))$
$=\Big[\hat{x}\frac{\partial}{\partial x}(f)+\hat{y}\frac{\partial}{\partial y}(f)+\hat{z}\frac{\partial}{\partial z}(f)\Big]+\Big[\hat{x}\frac{\partial}{\partial x}(g)+\hat{y}\frac{\partial}{\partial y}(g)+\hat{z}\frac{\partial}{\partial z}(g)\Big]$
$\Rightarrow\vec{\nabla{(fg)}}=f\vec{\nabla g}+g\vec{\nabla f}$
Hence proved
Let $\vec{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$ and $\vec{B}=B_x\hat{x}+B_y\hat{y}+B_z\hat{z}$
b)
$\vec{A}\times\vec{B}=(A_yB_z-A_zB_y)\hat{x}+(A_zB_x-A_xB_z)\hat{y}+(A_xB_y-A_yB_x)\hat{z}$
$\Rightarrow\vec{\nabla}\cdot(\vec{A}\times\vec{B})=\frac{\partial}{\partial x}(A_yB_z-A_zB_y)+\frac{\partial}{\partial y}(A_zB_x-A_xB_z)+\frac{\partial}{\partial z}(A_xB_y-A_yB_x)$
$=\Big[A_y\frac{\partial}{\partial x}B_z+B_z\frac{\partial}{\partial x}A_y-A_z\frac{\partial}{\partial x}B_y+B_y\frac{\partial}{\partial x}A_z\Big]+\Big[A_z\frac{\partial}{\partial y}B_x+B_x\frac{\partial}{\partial y}A_z-A_x\frac{\partial}{\partial y}B_z+B_z\frac{\partial}{\partial y}A_x\Big]+\Big[A_x\frac{\partial}{\partial z}B_y+B_y\frac{\partial}{\partial z}A_x-A_y\frac{\partial}{\partial z}B_x+B_x\frac{\partial}{\partial z}A_y\Big]$
$=B_x\Big(\frac{\partial}{\partial y}A_z-\frac{\partial}{\partial z}A_y\Big)+B_y\Big(\frac{\partial}{\partial z}A_x-\frac{\partial}{\partial x}A_z\Big)+B_z\Big(\frac{\partial}{\partial x}A_y-\frac{\partial}{\partial y}A_x\Big)-A_x\Big(\frac{\partial}{\partial y}B_z-\frac{\partial}{\partial z}B_y\Big)-A_y\Big(\frac{\partial}{\partial z}B_x-\frac{\partial}{\partial x}B_z\Big)-A_z\Big(\frac{\partial}{\partial x}B_y-\frac{\partial}{\partial y}B_x\Big)$
$=\vec{B}\cdot(\vec{\nabla}\times\vec{A})-\vec{A}\cdot(\vec{\nabla}\times\vec{B})$
$\therefore)\vec{\nabla}\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\vec{\nabla}\times\vec{A})-\vec{A}\cdot(\vec{\nabla}\times\vec{B})$
Hence proved
c)$\vec{\nabla}\times(f\vec{A})=\hat{x}\Big(\frac{\partial}{\partial y}(fA_z)-\frac{\partial}{\partial z}(fA_y)\Big)+\hat{y}\Big(\frac{\partial}{\partial z}(fA_x)-\frac{\partial}{\partial x}(fA_z)\Big)+\hat{z}\Big(\frac{\partial}{\partial x}(fA_y)-\frac{\partial}{\partial y}(fA_x)\Big)$
$=\Big[f\frac{\partial}{\partial y}A_z+A_z\frac{\partial}{\partial y}f-f\frac{\partial}{\partial z}A_y-A_y\frac{\partial}{\partial z}f\Big]\hat{x}+\Big[f\frac{\partial}{\partial z}A_x+A_x\frac{\partial}{\partial z}f-f\frac{\partial}{\partial x}A_z-A_z\frac{\partial}{\partial x}f\Big]\hat{y}+\Big[f\frac{\partial}{\partial x}A_y+A_y\frac{\partial}{\partial x}f-f\frac{\partial}{\partial y}A_x-A_x\frac{\partial}{\partial y}f\Big]\hat{z}$
$=\Big[f\frac{\partial}{\partial y}A_z-f\frac{\partial}{\partial z}A_y\Big]\hat{x}+\Big[f\frac{\partial}{\partial z}A_x-f\frac{\partial}{\partial x}A_z\Big]\hat{y}+\Big[f\frac{\partial}{\partial x}A_y-f\frac{\partial}{\partial y}A_x\Big]\hat{z}-\Big[(A_y\frac{\partial}{\partial z}f-A_z\frac{\partial}{\partial y}f)\hat{x}+(A_z\frac{\partial}{\partial x}f-A_x\frac{\partial}{\partial z}f)\hat{y}+(A_x\frac{\partial}{\partial y}f-A_y\frac{\partial}{\partial x}f)\hat{z}\Big]$
$=f(\vec{\nabla}\times\vec{A})-\vec{A}\times(\vec{\nabla f})$
$\therefore\vec{\nabla}\times(f\vec{A})=f(\vec{\nabla}\times\vec{A})-\vec{A}\times(\vec{\nabla f})$
Hence proved
