We can begin by rewriting the equation for a circle as

$\displaystyle 3x^2 + 9y^2 = 27 \Rightarrow \left(\frac{x}{3} \right )^2 + \left(\frac{y}{\sqrt{3}} \right )^2=1$.

This directly tells us that $\displaystyle x = 3 \cos t, y = \sqrt{3} \sin t$.  This allows us to write our final expression for the parametric representation as

$\displaystyle \vec{r}(t)= \left< 3 \cos t, \sqrt{3} \sin t \right>$

The conversion from Cartesian to cylindrical coordinates is as follows:

$\displaystyle x=r\cos \theta$

$\displaystyle y=r \sin \theta$

$\displaystyle z=z$

The three components of the vector then become:

$\displaystyle 3xy\rightarrow3r\cos \theta \:r\sin \theta=3r^2\cos\theta \sin\theta$

$\displaystyle x^2z\rightarrow r^2 \cos^2\theta z$

$\displaystyle y\rightarrow r\sin \theta$

The coordinates (2, 1, -2) corresponds to: x = 2, y = 1, z = -2, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\displaystyle \\ r = \sqrt{x^{2}+y^{2}} \\ \tan(\theta) = \frac{y}{x} \\ z = z$

So, filling in for x, y, z:

$\displaystyle \\ r = {\sqrt{2^{2}+1^{2}}} = \sqrt{5} \\ \tan(\theta) = \frac{1}{2} \Rightarrow \theta = \tan^{-1}\frac{1}{2} \\ z = -2$

Then the cylindrical coordinates are represented as:

$\displaystyle \left ( \sqrt{5}, \tan^{-1}\frac{1}{2}, -2 \right )$

The coordinates (0, 3, 4) corresponds to: x = 0, y = 3, z = 4, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\displaystyle \\ r = \sqrt{x^{2}+y^{2}} \\ \tan(\theta) = \frac{y}{x} \\ z = z$

So, filling in for x, y, z:

$\displaystyle \\ r = {\sqrt{0^{2}+3^{2}}} = \sqrt{9} = 3 \\ \tan(\theta) = \frac{3}{0} \Rightarrow \theta = \frac{1}{2}\pi \\ z = 4$

Then the cylindrical coordinates are represented as:

$\displaystyle \left (3, \frac{1}{2}\pi, 4 \right )$

The coordinates (√2, 1, 1) corresponds to: x = √2, y = 1, z = 1, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\displaystyle \\ r = \sqrt{x^{2}+y^{2}} \\ \tan(\theta) = \frac{y}{x} \\ z = z$

So, filling in for x, y, z:

$\displaystyle \\ r = {\sqrt{\sqrt{2}^{2}+1^{2}}} = \sqrt{2+1} = \sqrt{3} \\ \tan(\theta) = \frac{1}{\sqrt{2}} \Rightarrow \theta = \tan^{-1}\frac{\sqrt{2}}{2} \\ z = 1$

Then the cylindrical coordinates are represented as:

$\displaystyle \left (\sqrt{3}, \tan^{-1}\frac{\sqrt{2}}{2}, 1 \right )$

The coordinates (3, π/3, -4) corresponds to: r = 3, θ = π/3, z = -4, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\displaystyle \\ x = r\cos\theta \\ y = r\sin\theta \\ z = z$

So, filling in for r, θ, z:

$\displaystyle \\ x = 3\cos{\frac{1}{3}\pi} = 3\cdot \frac{1}{2} = \frac{3}{2} \\ y = 3\sin{\frac{1}{3}\pi} = 3\cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{2}}{2} \\ z = -4$

Then the Cartesian coordinates are represented as:

$\displaystyle \left (\frac{3}{2}, \frac{3\sqrt{3}}{2}, -4 \right)$

The coordinates (-2, 2, 3) corresponds to: x = -2, y = 2, z = 3, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\displaystyle \\ r = \sqrt{x^{2}+y^{2}} \\ \tan(\theta) = \frac{y}{x} \\ z = z$

So, filling in for x, y, z:

$\displaystyle \\ r = {\sqrt{\left ( -2\right )^{2}+2^{2}}} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \\ \tan(\theta) = \frac{2}{-2} \Rightarrow \theta = \tan^{-1}\ \left ( -1\right ) \Rightarrow \theta = \frac{3}{4}\pi \\ z = 3$

Then the cylindrical coordinates are represented as:

$\displaystyle \left ( 2\sqrt{2}, \frac{3}{4}\pi, 3\right )$

The coordinates (1, 45°, 1) corresponds to: r = 1, θ = 45°, z = 1, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\displaystyle \\ x = r\cos\theta \\ y = r\sin\theta \\ z = z$

So, filling in for r, θ, z:

$\displaystyle \\ x = 1\cos{45^{\circ}} = 1\cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ y = 1\sin{45^{\circ}} = 1\cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ z = 1$

Then the Cartesian coordinates are represented as:

$\displaystyle \left ( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}, 1\right )$

The coordinates (2, π/2, -4) corresponds to: r = 2, θ = π/2, z = -4, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\displaystyle \\ x = r\cos\theta \\ y = r\sin\theta \\ z = z$

So, filling in for r, θ, z:

$\displaystyle \\ x = 2\cos{\frac{1}{2}\pi} = 2\cdot 0 = 0 \\ y = 2\sin{\frac{1}{2}\pi} = 2\cdot 1 = 2 \\ z = -4$

Then the Cartesian coordinates are represented as:

$\displaystyle \left ( 0,2,-4\right )$

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\displaystyle \rho=\sqrt{x^2+y^2+z^2}$

$\displaystyle \theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\displaystyle \rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$

$\displaystyle \theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$