You are incorrect about the first problem, I think. It doesn't say 'for all y', it says 'there exists a y'. So for all x, there is some y such that for all z, x+y=z. Your choice of 1 for y was arbitrary, and not warranted given that it just says 'there exists a y', not that any y you choose will satisfy the condition. It's for all x and z, so let's choose 2 random numbers, say 3 and 34, for x and z. Now, is there some y that when added to 3 makes 34? Of course there is: 31. And that will hold true no matter what numbers you choose for x and z.
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).