I'm trying to do this problem and I feel like I've made some progress but I've become quite stuck
Suppose that $h_∗$ is an ordinary homology theory with coefficients the abelian group $G$. Calculate $\overline{h}∗(\mathbb CP^2)$ and $\bar{h}∗(\sum \mathbb CP^2)$.
My attempt:
$\pi_3(S^2) \cong \mathbb Z$. Let $\phi: S^3 \rightarrow S^2$ be the generator $+1 \in \mathbb Z$. Then $\mathbb CP^2$ is the homotopy cofibre of $\phi$.
$\pi_4(S^3) \cong \mathbb Z/2\mathbb Z$ and $\sum \phi: S^4 \rightarrow S^3$ is a generator for this group. The homotopy cofibre of $\sum \phi$ is $\sum \mathbb CP^2$
$\pi_5(S^3) \cong \mathbb Z/2\mathbb Z$ and a generator of this group is represented by the composite $\epsilon: S^5 \stackrel{\sum^2\phi}\rightarrow S^4 \stackrel{\sum\phi}\rightarrow S^3$.
From here I am stumped so any help would be much appreciated!
The easy way to proceed is to make use of the (homological) Universal Coefficient Theorem. Since we know very well the homology of $\Bbb CP^2$, a quick application of it yield that $H_k(\Bbb CP^2;G)\cong G$ if $k$ even and $k\leq 4$, $0$ otherwise. In fact the short exact sequence looks like: $$0 \to H_{k-1}(\Bbb CP^2)\otimes G \to H_k(\Bbb CP^2;G) \to Tor(H_k(\Bbb CP^2),G)\to 0$$ for $k$ even we have that $k-1$ is odd and therefore $H_{k-1}(\Bbb CP^2)=0$ so $$0 \to \Bbb Z \otimes G \to H_k(\Bbb CP^2;G) \to Tor(0,G)=0\to 0$$ gives the result.
$k$ odd is done similarly.
Reduced homology behave like you would expect, so the only difference is that the $0$-th homology group is $0$.
If you don't want to make use of UCT, there is cellular homology with coefficient, which gives the (same) result with the same reasoning you would do for the integer coefficient.
For what concern the suspension, well make use of the Suspension isomorphism $\tilde{H}_k(X;G) \cong \tilde{H}_{k+1}(\Sigma X;G)$ to conclude
Such isomorphism is nothing fancy, you can obtain it via a Mayer-Vietoris sequence applie to $\Sigma X$ and its cover given by the two cones.