Your method is correct. You can check your own solutions to verify that they are correct.
35*6 = 210 which is congruent to 10 mod 50.
35*16 = 560 which is also congruent to 10 mod 50.
Similarly for the other solutions that you found.
You can verify that 22, 28, and 40 are not solutions.
35*22...
For the first part, you might want to show the step between your use of the definition and your conclusion. For example:
b ≥ 0 => -b ≤ 0 => -b ≤ 0 ≤ b. Now assume |a| ≤ b, if a ≥ 0 then |a| = a and we have 0 ≤ |a| = a ≤ b => -b ≤ 0 ≤ a ≤ b => -b ≤ a ≤ b, which is what was desired...
I think there is some more to it than that. What happens if f(a) = 0? Remember f is mapping to the non-negative integers, so you need to consider this case. This will help you complete the proof, I think.
This one is tricky.
Instead of trying to prove directly that A is closed, try to show that the complement, X-A, is open. Show the set of all x such that f(x) > g(x) to be open. You can show this by showing that every point x in X is an interior point. Just find an open set around an...
If V = <(1,1)> then how can (1,0) be in V? There is no scalar a such that a*(1,1) = (1,0). Similarly, there is no scalar b such that b*(1,0) = (1,1). So the intersection of U and V is indeed (0,0).
Take a look at what M is, when written in the form RxS for the sets R and S. You will see that both R and S are well known sets and that should help you figure out what ZxZ/M should look like.
I assume that p is fixed in M?
Do you know what Z/pZ is isomorphic to, for a given prime p?
Do you know how to show that ZxZ / Zx{0} is isomorphic to Z?
This depends on your precise definition of maximum. If you assume S has a maximum, then what do you know about it?
Similarly, if S is bounded above, and the sup S, call it s0, is an element inside of S, then what do we know about s0?