To convert polar coordinates $\displaystyle (r, \theta)$ to rectangular coordinates $\displaystyle (x, y)$,

$\displaystyle x=r\cos\theta$

$\displaystyle y=r\sin\theta$

Using the information given in the question, 

$\displaystyle x=5\cos \frac{3\pi}{2}=5(0)=0$

$\displaystyle y=5 \sin \frac{3\pi}{2}=5(-1)=-5$

The rectangular coordinates are $\displaystyle (0, -5)$

Start by multiplying both sides by $\displaystyle r$.

$\displaystyle r=4\cos\theta$

$\displaystyle r^2=4r\cos\theta$

Keep in mind that $\displaystyle r^2=x^2+y^2$

$\displaystyle x^2+y^2=4r\cos\theta$

Remember that $\displaystyle x=r\cos\theta$

So then,

$\displaystyle x^2+y^2=4x$

$\displaystyle x^2-4x+y^2=0$

Now, complete the square.

$\displaystyle (x^2-4x+4)+y^2=4$

$\displaystyle (x-2)^2+y^2=4$

Start by multiplying both sides by $\displaystyle r$.

$\displaystyle r=8\sin\theta$

$\displaystyle r^2=8r\sin\theta$

Remember that $\displaystyle r^2=x^2+y^2$

$\displaystyle x^2+y^2=8r\sin\theta$

Keep in mind that $\displaystyle y=r\sin\theta$

So then,

$\displaystyle x^2+y^2=8y$

$\displaystyle x^2+y^2-8x=0$

Now, complete the square.

$\displaystyle x^2+(y^2-8x+16)=16$

$\displaystyle x^2+(y-4)^2=16$

This is a graph of a circle with a radius of $\displaystyle 4$ and a center at $\displaystyle (0, 4)$

Remember that $\displaystyle \sec\theta=\frac{1}{\cos\theta}$

So then $\displaystyle r=8\sec\theta$ becomes

$\displaystyle r=\frac{8}{\cos\theta}$

Now, multiply both sides by $\displaystyle \cos\theta$ to get rid of the fraction.

$\displaystyle r\cos\theta=8$

Since $\displaystyle x=r\cos\theta,$ the rectangular form of this equation is

$\displaystyle x=8$

Start by using the double angle formula for $\displaystyle \sin$.

$\displaystyle \sin(2\theta)=2\sin\theta\cos\theta$

Substitute that into the equation gives the following:

$\displaystyle r^2=4(2\sin\theta\cos\theta)$

$\displaystyle r^2=8\sin\theta\cos\theta$

Because we need $\displaystyle r\cos\theta$ and $\displaystyle r\sin\theta$ to get $\displaystyle x$ and $\displaystyle y$ respectively, multiply both sides by $\displaystyle r^2$.

$\displaystyle (r^2)(r^2)=8(r\sin\theta)(r\cos\theta)$

Now, recall that $\displaystyle r^2=x^2+y^2$.

$\displaystyle (x^2+y^2)(x^2+y^2)=8xy$

$\displaystyle (x^2+y^2)^2=8xy$

Start by taking the tangent.

$\displaystyle \tan\theta=\tan\frac{5\pi}{6}$

$\displaystyle \tan\theta=-\frac{\sqrt3}{3}$

Recall that $\displaystyle \tan\theta=\frac{y}{x}$

$\displaystyle \frac{y}{x}=-\frac{\sqrt3}{3}$

$\displaystyle y=-\frac{\sqrt3}{3}x$

Start by taking the tangent.

$\displaystyle \tan\theta=\tan\frac{5\pi}{4}$

$\displaystyle \tan\theta=1$

Recall that $\displaystyle \tan\theta=\frac{y}{x}$

$\displaystyle \frac{y}{x}=1$

$\displaystyle y=x$

Recall that $\displaystyle \sec\theta=\frac{1}{\cos\theta}$

Substituting this into the given equation gives

$\displaystyle r=4\frac{1}{\cos\theta}\tan\theta$

Multiply both sides by $\displaystyle \cos\theta$ to get rid of the fraction.

$\displaystyle r\cos\theta=4\tan\theta$

Recall that $\displaystyle x=r\cos\theta$ and that $\displaystyle \tan\theta=\frac{y}{x}$

$\displaystyle x=4\frac{y}{x}$

$\displaystyle 4y=x^2$

$\displaystyle y=\frac{1}{4}x^2$

Recall that $\displaystyle \csc\theta=\frac{1}{sin\theta}$ and $\displaystyle \cot\theta=\frac{1}{tan\theta}$

$\displaystyle r=2\frac{1}{\sin\theta}\frac{1}{\tan\theta}$

Multiply both sides by $\displaystyle \sin\theta$

$\displaystyle r\sin\theta=2(\frac{1}{\tan\theta})$

Recall that $\displaystyle \tan\theta=\frac{y}{x}$ and $\displaystyle r\sin\theta=y$

$\displaystyle y=2(\frac{1}{\frac{y}{x}})$

$\displaystyle y=2(\frac{x}{y})$

$\displaystyle 2x=y^2$

$\displaystyle y=\sqrt{2x}$

Recall that $\displaystyle \sec\theta=\frac{1}{\cos\theta}$

Plugging this into the equation gives us

$\displaystyle r=-3\frac{1}{\cos\theta}$

Multiply both sides by $\displaystyle \cos\theta$ to get rid of the fraction.

$\displaystyle r\cos\theta=-3$

Recall that $\displaystyle r\cos\theta=x$

So then the rectangular form of the equation is $\displaystyle x=-3$