For (b):
Note that $$h = g\circ f(x,y) = g(x+y, x) = (y, x).$$
For (c):
Surjective? No. For this to be true you would have to be able to find $x,y$ such that $f(x,y) = (x + y, x)= (1,2)$. That would mean $x = 2$ and $y = -1 \notin \mathbb{N}$. So that is not possible.
Injective? Say that $f(x,y) = f(x', y')$. We are assuming that two different inputs give the same output. For $f$ to be injective we need to prove that the inputs actually are the same. So we have $f(x,y) = f(x',y')$ and we need to prove that $x= x'$ adn $y = y'$. That $f(x,y) = f(x', y')$ means that $(x+y, x) = (x' + y', x')$. But if this is true then we certainly have that $x = x'$. And if $x=x'$ and $x + y = x' + y'$, then it follows that $y = y'$. Hence $f$ is injective. Note that $x'$ (the backtick or prime or what ever one calls it) is just another element. It isn't doesn't mean that it is necessarily related tot $x$. One could have chosen another variable name instead.
For (d):
Now say that $(n,m)\in \mathbb{N} \times \mathbb{N}$ with $n\geq m$. To prove that $f$ now is surjective, you want to find $x,y$ such that $$f(x,y) = (n,m)$$. But since $n\geq m$ you can write $n = m + a$ for $a\geq 0$. So you want $(x,y)$ such that
$$x + y = n\quad \text{and}\quad x = m.$$ Well pick $x = m$. Then left is to find $y$ such that $x+y = n$, and with the choice of $x$, that means that we want $y$ so that $m + y = n$. That is $m + y = m + a$. This you have if you exactly if pick $y = a$. So $f$ is both injective and surjective here. It is clearly surjective and the injectivity follows from the fat that there was only one way to pick $x$ and $y$.
(I guess that there is the subtle assumption that $0\in \mathbb{N}$.)
For your first question, suppose that we had two elements of $A$ with the same coordinate: that is, we had $(a, b) \in A$ and $(a, c) \in A$. Then by definition of $A$, we have that $b^2 = a = c^2$, so $b^2 = c^2$. So does it follow that $b = c$? (It doesn't.)
For the second, you are correct that it is injective: For if $f(n) = f(m)$, then $$(2n, n + 3) = (2m, m + 3) \implies n + 3 = m + 3 \implies n = m$$ You're also correct that it's not onto, since it never hits $(1, 0)$. If it were onto, then you'd select an arbitrary element of $\Bbb{Z} \times \Bbb{Z}$ and find an element of $\Bbb{Z}$ mapping to it.
Best Answer
By definition, $f(x) \geq 0$, so your second counterexample is correct. So is your first,for obvious reasons. But it's interesting to note if we redefine the domain and range as follows: $f:\mathbb{R^{+}} -> \mathbb{R^{+}}$ where $R^{+} =\{x\in R | x\geq 0\}$ or $f:\mathbb{R^{-}} -> \mathbb{R^{-}}$ where $R^{-} =\{x\in R | x\leq 0\}$ . Then it's easy to show this is a bijection.
But no, the ordinary absolute value function isn't a bijection. The much tricker thing to prove is whether or not it's continuous on R. Can you prove it is and if not, give a counterexample?
Ain't math fun?