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 $\operatorname{Tor}$...)
You can find the construction of homology with general coefficients and the universal coefficient theorem in Hatcher's Algebraic Topology, which is available free from his website.
The answer to your third question is yes.
The answer to the second part of your first question is yes, especially in the case that we take $G$ to be a field, most often finite or $\mathbb{Q}$, or $\mathbb{R}$ in differential topology. Homology over a field is simple because $\operatorname{Tor}$ always vanishes, so you get e.g. an exact duality between homology and cohomology. Homology with $\mathbb{Z}_2$ coefficients is also the appropriate theory for many questions about non-orientable manifolds-their top $\Bbb{Z}$-homology is zero, but their top $\Bbb{Z}_2$ homology is $\Bbb{Z}_2$, which leads to the degree theory in Milnor you were mentioning.
Cohomology with more general coefficients than $\mathbb{Z}$ is even more useful than homology. For instance it leads to the result that if a manifold $M$ has any Betti number $b_i(M)<b_i(N)$, where $b_i$ is the rank of the free part of $H_i$, there's no map $M\to N$ of non-zero degree. This has lots of quick corollaries-for instance, there's no surjection of $S^n$ onto any $n$-manifold with nontrivial lower homology! Edit: This is obviously false, and I no longer have any idea whether I meant anything true.
But in the end $H_*(X;G)$ is more of a stepping stone than anything else; it gets you thinking about how much variety there could be in theories satisfying the axioms of homology. It turns out there's almost none-singular homology with coefficients in $G$ is the only example-but if we rid ourselves of the "dimension axiom"
$$H_*(\star)=\left\{\begin{matrix}\mathbb{Z},*=0\\0,*>0\end{matrix}\right.$$
then we get a vast collection of "generalized (co)homology theories," beginning with K-theory, cobordism, and stable homotopy, which really do contain new information. In some cases, so much new information that we can't actually compute them yet!
Best Answer
That's a great question ! And in the beginning, examples aren't easy to find.
For the first one, there are things such as (Poincaré) homology spheres : those are spaces which have the homology of a sphere, but a nontrivial $\pi_1$. Depending on what you already know, this may or may not be easy to construct, so here's a more elementary example : take $\mathbb T = S^1 \times S^1$ on one side and $S^1\vee S^1 \vee S^2$ on the other side. It's easy to see that their homologies are the same, but their $\pi_1$ and $\pi_2$ (and actually many others) differ.
A way to understand why examples are not too easy to find : any map between simply-connected spaces which induces an isomorphism on $H_*$ also induces one on $\pi_*$. Watch out : this is not saying that simply-connected spaces with the same $H_*$ have the same $\pi_1$ : there really has to be a map. Can you tweak my example to see why that is ?
As for spaces with the same $\pi_*$ but not homeomorphic , this is easy : just take any non-singleton contractible space (such as $\mathbb R$) and more generally homotopy-equivalent spaces that aren't homeomorphic. So maybe you meant "same $\pi_*$ but not homotopy equivalent" Depending on what you know, this can also be tricky, for two reasons : the first is that again any map (between nice spaces) inducing an isomorphism on $\pi_*$ is a homotopy equivalence (note that once again we require a map !). The second reason is that, without a map, any two (nice) spaces with exactly one nonzero homotopy group, which are isomorphic, (say $\pi_n(X) \cong \pi_n(Y)$ for all $n$, and all of them are zero except for one $k$) are homotopy equivalent (those are called Eilenberg-MacLane spaces).
So with that in mind, it can be hard to find examples. If you know about the long exact sequence of a fibration, then here's one example : $S^3\times \mathbb CP^\infty$ and $S^2$ have isomorphic homotopy groups, but are not homotopy equivalent
(To see it, use the long exact sequence of the Hopf fibration $S^1\to S^3\to S^2$)
I don't know a simpler example, however, so it really depends on what you know.