Answer
\[\begin{align}
& \left( \mathbf{a} \right)\left( 4,\frac{3\pi }{4},2 \right) \\
& \left( \mathbf{b} \right)\left( 2\sqrt{5},\frac{3\pi }{4},\arccos \left( \frac{1}{\sqrt{5}} \right) \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \left( -2\sqrt{2},2\sqrt{2},2 \right) \\
& \left( \mathbf{a} \right)\text{We have the rectangular coordinates }\left( -2\sqrt{2},2\sqrt{2},2 \right) \\
& \left( x,y,z \right):\text{ }\left( -2\sqrt{2},2\sqrt{2},2 \right) \\
& \text{Rectangular to cylindrical} \\
& {{r}^{2}}={{x}^{2}}+{{y}^{2}}\to r=\sqrt{{{\left( -2\sqrt{2} \right)}^{2}}+{{\left( 2\sqrt{2} \right)}^{2}}}=4 \\
& \tan \theta =\frac{y}{x}\to \theta ={{\tan }^{-1}}\left( \frac{2\sqrt{2}}{-2\sqrt{2}} \right)=\frac{3\pi }{4} \\
& z=z\to z=2 \\
& \text{The cylindrical coordinates are:} \\
& \left( 4,\frac{3\pi }{4},2 \right) \\
& \\
& \left( \mathbf{b} \right)\text{We have the rectangular coordinates }\left( -2\sqrt{2},2\sqrt{2},2 \right) \\
& \left( x,y,z \right):\left( -2\sqrt{2},2\sqrt{2},2 \right) \\
& \text{Rectangular to spherical }\left( \rho ,\theta ,\phi \right) \\
& {{\rho }^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}\to \rho =\sqrt{{{\left( -2\sqrt{2} \right)}^{2}}+{{\left( 2\sqrt{2} \right)}^{2}}+{{\left( 2 \right)}^{2}}}=2\sqrt{5} \\
& \tan \theta =\frac{y}{x}\to \theta ={{\tan }^{-1}}\left( \frac{2\sqrt{2}}{-2\sqrt{2}} \right)=\frac{3\pi }{4} \\
& \phi =\arccos \left( \frac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \right)\to \phi =\arccos \left( \frac{2}{2\sqrt{5}} \right) \\
& \phi =\arccos \left( \frac{1}{\sqrt{5}} \right) \\
& \text{The spherical }\left( \rho ,\theta ,\phi \right)\text{ coordinates are:} \\
& \left( 2\sqrt{5},\frac{3\pi }{4},\arccos \left( \frac{1}{\sqrt{5}} \right) \right) \\
\end{align}\]
