To solve the first one, it's best to change the problem to Cartesian first, just to see it much easier. In order to do that, you'll want to do this:

*multiply both sides by r*

*substitute for *

*substitute in for r*

Now you have your Cartesian. So, now let's solve for x, much easier than to solve for y. You should get

Now, since we want the inverse, just look at the Domain of the above function and you have the Range you wanted originally, which is

to answer part 2, we look at first and see that for is negative when \pi/2,3\pi/4)" alt="\Theta\pi/2,3\pi/4)" />, but since it's squared, then we don't have to worry about that term. So, since is positive for all and since the largest value of is 1, then 2-1 is 1, hence positive.

Hope this helps.