First of all, I suggest you find a good article/book on Cantor's work on multiple infinities. Reading that will enlighten you and answer probably most of the questions you have concerning this subject. Try these lecture notes.
But to give a brief answer to your current questions:
Per definition, two sets have the same cardinality if there exists a bijection between them. This definition was introduced by Cantor because for infinite sets, you could not just count the elements. Not all infinite sets have the same cardinality. For example, the natural numbers $\mathbb{N}$ and the real numbers $\mathbb{R}$ do not have the same cardinality. This was proven with the diagonal argument.
$\mathbb{N}$ and $\mathbb{Z}$ are both infinite sets. I suggest you check out their definitions on wikipedia (the natural numbers, the integers). They also, like Ali said in his post, have the same cardinality. The bijection that was given by him is probably the easiest one to grasp.
As for $\mathbb{Q}$, the rational numbers, this is also an infinite set that has the same cardinality as $\mathbb{N}$ (and thus also $\mathbb{Z}$). I suggest you check out the bijection that is given in the lectures notes that I linked above, that one is probably the clearest one.
EDIT: I noticed that your book is not using the usual set-theoretic definition of natural numbers, where $n = \{0,1,2,\ldots, n-1\}$. Namely, $0$ is not considered to be a natural number, so the prototypical $n$-element set is the set $\{1,\ldots,n\}$ rather than $n$ itself. I will update my answer to reflect this.
I think you are making this more complicated that it needs to be. The notion of countability does not seem to be relevant. As you said, we define a set $x$ to be finite if there is a bijection between $x$ and some natural number $n \in \mathbb{N}$ (or it is empty.) Let $x$ be a nonempty set.
If $x$ is finite, then there is a bijection from $x$ onto a finite set $y$: namely, let $y=x$ and consider the identity function.
Conversely, assume there is a bijection $f$ from $x$ onto a finite set $y$. By definition of "finite" there is a bijection $g$ from $y$ onto the set $\{1,\ldots,n\}$ for some natural number $n$. You can then show that the composition $g \circ f$ is a bijection from $x$ onto the set $\{1,\ldots,n\}$, so $x$ is finite as well.
One problem with your argument for the forward direction is that the theorem you are applying deals with countable sets rather than finite sets, so you cannot use it to conclude that there is a bijection from anything to a finite set. In fact, the theorem is irrelevant. I didn't finish reading the rest of this direction.
Your argument for the reverse direction is vague. You are proving that something is finite, so you should mention some natural number $n$, some set of the form $\{1,\ldots,n\}$, and some bijection between this set and some other set.
Best Answer
Suppose $A$ is bijective to $[n] := \{1, 2, \ldots, n\}$ for some $n$. Then any proper subset $\tilde{A}$ of $A$ will have cardinality $|\tilde{A}|= m$ for some $m<n$ (in other words, $\tilde{A}$ is bijective to $[m]$ with $m<n$). By the hint given at the end of the problem, no map from $[n]$ to $[m]$ is injective, so $[n]$ is not bijective to $[m]$. Hence, $A$ is not bijective with $\tilde{A}$, for any proper subset $\tilde{A}$. By definition, $A$ is finite.
I'll leave it to you to prove the other direction.