It turns out that the homology groups $H_{\bullet}(X,\mathbb{Z})$ of a space $X$ with $\mathbb{Z}$-coefficients determine the homology groups $H_{\bullet}(X,A)$ with coefficients in any abelian group $A$. The key result here is the (very well named!) Universal Coefficient Theorem.
The basic idea is that our first guess at $H_i(X,A)$ is simply $H_i(X,\mathbb{Z}) \otimes A$. This is a good first guess in that in all cases there is a natural injective map $H_i(X,\mathbb{Z}) \otimes A \rightarrow H_i(X,A)$. As you have seen, this map need not be an isomorphism. The universal coefficient theorem tells you that its cokernel is $\operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A)$ and also that the sequence is (non-canonically) split, i.e.,
$$H_i(X,A) \cong (H_i(X,\mathbb{Z}) \otimes A) \oplus \operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A).$$
Here $\operatorname{Tor}( \ , \ )$ is the first "Tor group" of homological algebra. It may well be that you don't know what this gadget is. (I didn't when I first learned algebraic topology.) So I found it helpful to write down a "cheatsheet" for $\operatorname{Tor}(X,Y)$ when $X$ and $Y$ are both finitely generated abelian groups. Indeed, since $\operatorname{Tor}$ is bi-additive and symmetric, it is enough to know that for all $m,n \in \mathbb{Z}^+$,
$\operatorname{Tor}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) = 0$ and
$\operatorname{Tor}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/\operatorname{gcd}(m,n) \mathbb{Z}$.
As a first exercise, try to use all this information to confirm that the homology groups of $\mathbb{R} \mathbb{P}^2$ with $\mathbb{Z}/2\mathbb{Z}$-coefficients are as you said.
(There is also a Universal Coefficient Theorem for cohomology in which the correction term involves $\operatorname{Ext} = \operatorname{Ext}^1$ instead of $\operatorname{Tor}$...)
Essentially, $$H_0(\mathbb Q;\mathbb Z)=\oplus_{q\in\mathbb Q} \mathbb Z$$ while $$H^0(\mathbb Q;\mathbb Z)=\prod_{q\in\mathbb Q}\mathbb Z$$
In particular, $H_0$ is countable, while $H^0$ is uncountable.
Best Answer
First note, that the only continuous maps $S^n \to \mathbb Q$ and $D^{n-1} \to \mathbb Q$ of the $n$-dimensional sphere and the $(n-1)$-dimensional Disk to the rationals are constant ($n\geq 1$). Therefore $\pi_k(\mathbb Q, *) = H_k(\mathbb Q, \mathbb Z) = H^k(\mathbb Q, \mathbb Z) = 0$ for all $k \geq 1$ and every basepoint $* \in \mathbb Q$. The statement on (co-)homology can easily be seen by considering singular (co-)homology.