x = r*cos(theta)

y = r*sin(theta)

r = sqrt(x^2 + y^2)

theta = tan^(-1)(y/x)

Two comments about the last equation. It's "theta" not "feta" and tan^(-1) is the inverse tangent function.

1) r^2 = x^2 + y^2 so,

r^2 = 4.

2) 0 <= t <= pi <== It's "pi", not "pie"

x=(cost) -2

y= (sin t) +3

Solve the x equation for cos(t):

x + 2 = cos(t)

Set up a right triangle. cos(t) is "adjacent" over hypotenuse so we want the side of the triangle adjecent to angle t to be x + 2 and the hypotenuse 1. Thus the "opposite" side will be sqrt{1 - (x + 2)^2}. So:

sin(t) = sqrt{1 - (x + 2)^2}/1 = sqrt{1 - (x + 2)^2}

So

y= (sin t) + 3 = sqrt{1 - (x + 2)^2} + 3

3) I can't do this one here. I'll leave it up to you.

4) r=2cos(theta) - 4sin(theta)

sqrt{x^2 + y^2} = 2*x/r - 4*y/r <== Square both sides

x^2 + y^2 = (2*x/r - 4*y/r)^2

x^2 + y^2 = 4x^2/r^2 + 16y^2/r^2 - 16xy/r^2

(x^2 + y^2)r^2 = 4x^2 + 16y^2 - 16xy

But r^2 = x^2 + y^2

(x^2 + y^2)^2 = 4(x^2 + 4y^2 - 4xy)

This will do. You can always solve this for y if you wish.

-Dan