Answer
$div\left( {{{\bf{e}}_r}} \right) = \frac{2}{{\sqrt {{x^2} + {y^2} + {z^2}} }}$
$curl\left( {\bf{F}} \right) = {\bf{0}}$
Work Step by Step
We have ${{\bf{e}}_r} = {r^{ - 1}}\left( {x,y,z} \right)$, where $r = \sqrt {{x^2} + {y^2} + {z^2}} $.
So,
${{\bf{e}}_r} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }},\frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }},\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)$
1. $div\left( {{{\bf{e}}_r}} \right) = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}$
$div\left( {{{\bf{e}}_r}} \right) = \frac{{{y^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + \frac{{{x^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + \frac{{{x^2} + {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}$
$div\left( {{{\bf{e}}_r}} \right) = \frac{{2{x^2} + 2{y^2} + 2{z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}$
$div\left( {{{\bf{e}}_r}} \right) = \frac{{2\left( {{x^2} + {y^2} + {z^2}} \right)}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} = \frac{2}{{\sqrt {{x^2} + {y^2} + {z^2}} }}$
2. $curl\left( {\bf{F}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}}&{\frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}}&{\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}}
\end{array}} \right|$
$curl\left( {\bf{F}} \right) = \left( { - \frac{{yz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + \frac{{yz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}} \right){\bf{i}}$
$ - \left( { - \frac{{xz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + \frac{{xz}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}} \right){\bf{j}}$
$ + \left( { - \frac{{xy}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}} + \frac{{xy}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}} \right){\bf{k}}$
$curl\left( {\bf{F}} \right) = {\bf{0}}$