Let $X$ be any set. Then, the cardinality of $\mathcal{P}(X)$ is $\left| 2^X \right| = 2^{|X|}$, since there is a bijection $\mathcal{P}(X) \to 2^X$, where $2^X$ is the set of all functions $X \to \{0, 1\}$. I'm afraid this is best answer that I can give, since, for example, it is not possible to prove that $\aleph_1 = | \mathcal{P}(\mathbb{N}) |$, or that $\aleph_1 \ne | \mathcal{P}(\mathbb{N}) |$, from the usual axioms of mathematics (ZFC).
In terms of aleph numbers, the cardinalities of $\mathbb{R}$, $\mathbb{Q}$, and $\mathcal{P}(\mathbb{Q})$ are $2^{\aleph_0}$, $\aleph_0$, and $2^{\aleph_0}$, respectively. $\aleph_0 = | \mathbb{N} |$ by definition. Here's an interesting fact: the set of all functions $\mathbb{R} \to \mathbb{R}$ has cardinality strictly greater than $| \mathbb{R} |$, but the subset of all continuous functions has cardinality equal to $| \mathbb{R} |$. (Can you prove this?)
Now, as for the "set" of all functions — this isn't a well-defined set under ZFC. So you can't talk about the cardinality. However, if you fix the domain and codomain, then we do get a set, and we write $Y^X$ for the set of all functions $X \to Y$. The cardinality, as the notation suggests, is $|Y|^{|X|}$. (Actually, this is a small lie. We define the notation $|Y|^{|X|}$ to mean $\left| Y^X \right|$. But it happens to agree with the arithmetic definition of exponentiation in the case where $X$ and $Y$ are finite sets.)
For each $a\in(0,1)$ we'll define a function $f(n) = a+n$. $f$ is clearly in $\mathbb{N}\to\mathbb{R}$ and since it's monotone it is also injective. Moreover, if $a_1\ne a_2$ then $f_1(n) \ne f_2(n)$.
Hence, the set of all functions described is subset of $A$ and it's cardinality is $\left|\left(0,1\right)\right| = \aleph$
Best Answer
One approach is to hypothesize some cardinal $\kappa$ for which you can prove $\kappa \leq \left|\mathbb{Z}^{\mathbb{Z}}\right|$ and $\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \kappa$.
Here's a hint: $$\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \left|(2^{\mathbb{Z}})^{\mathbb{Z}}\right| = 2^{|\mathbb{Z} \times \mathbb{Z}|}$$