Although your answer to the second question was correct, the way you went about it shows the same error. There exists an x such that for all y, there exists a z such that x+y=z. You shouldn't start with any random x; although it works out in this case, that's the wrong approach. It's for all y, so y is the one where you can just choose any number, and the rest must be able to conform to it. So if you pick a 3 for y, you get x + 3 = z. It is easy to see that there will always be some combination of x and z that will make this true. But it is misleading to say 'let's choose a value for x' when it's 'there exists an x', not 'for all x'.

Incidentally, I'm assuming that the domain is the rationals. If it is the reals, then I'm not sure that either of these is true, since I don't think there is a specific real number that when added to pi, for instance, will make 5 (unless you're allowed to say pi plus (5 minus pi), of course).