Answer
$$\left( {2\sqrt {10} ,\frac{\pi }{4},\frac{{2\sqrt 5 }}{5}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {2,2,4\sqrt 2 } \right) \cr
& \left( {x,y,z} \right):{\text{ }}\left( {2,2,4\sqrt 2 } \right) \to x = 2,{\text{ }}y = 2,{\text{ }}z = 4\sqrt 2 \cr
& {\text{Rectangular to spherical }}\left( {\rho ,\theta ,\phi } \right) \cr
& {\rho ^2} = {x^2} + {y^2} + {z^2} \to \rho = \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {4\sqrt 2 } \right)}^2}} = 2\sqrt {10} \cr
& \tan \theta = \frac{y}{x} \to \theta = {\tan ^{ - 1}}\left( {\frac{2}{2}} \right) = \frac{\pi }{4} \cr
& \phi = \arccos \left( {\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) \to \phi = \arccos \left( {\frac{{4\sqrt 2 }}{{2\sqrt {10} }}} \right) \cr
& \phi = \arccos \left( {\frac{{2\sqrt 5 }}{5}} \right) \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {2\sqrt {10} ,\frac{\pi }{4},\frac{{2\sqrt 5 }}{5}} \right) \cr} $$
