My solution for #3:

Define by . We know that is continuous because and the identity map both are, and by the limit laws. Applying the intermediate value theorem (problem #2) we see that there exists such that . But this means , so we are done.

Counterexample for the open interval: consider defined by . First, we can verify that if then , so indeed maps to . To see that there is no fixed point, note that the only solution to in is , which is not in . (We can also view this graphically by plotting both and and checking that they do not intersect in .)

I took the survey.