Basic questions on the homotopy category
For products and coproducts, this is covered on page 67-68 of "Introduction to Homotopy Theory" by Martin Arkowitz
To see that the homotopy category is not complete or cocomplete (i.e. that there are diagrams which don't have limits or colimits), check out "Modern Classical Homotopy Theory" by Jeffrey Strom, page 435. He explicitly constructs a diagram with no colimit.
This question of yours is a good question, because thinking this way (i.e. asking for good categorical properties) has led to lots of theory over the years. I discuss two such cases below, although for your application they will probably not be helpful.
I study model categories, which are categories where one can "do homotopy theory." So $Top$ and $Top_\ast$ are examples. Let $M$ be a model category (feel free to think of $M$ as $Top$) and let $Ho(M)$ be its homotopy category. It turns out that you get arbitrary products and coproducts in $Ho(M)$ for any model category $M$ (see Hovey's book "Model Categories" 1.3.11). Furthermore, you can sometimes compute products on the model category level, as we'll now discuss.
If one wants $F:M\to Ho(M)$ to preserve products, then one wants $M$ to be a monoidal model category, i.e. to have the structure of a monoidal category (for $Top$ this means having a notion of product, which it does) and to satisfy a coherence condition which says that the monoidal products plays nicely with the model structure. It's proven in Mark Hovey's book Model Categories that $Top$ and $Top_\ast$ are monoidal model categories, i.e. have that coherence condition. So $Ho(Top)$ is a monoidal category whose unit is the image under $F$ of the unit in $Top$. In $Top$ the unit is $\ast$ and so it must be $\ast$ again in $Ho(Top)$. Furthermore, this means you can compute products on the model category level, so if someone gives you two objects of $Ho(Top)$ you can find an object $X\in Top$ such that $F(X)$ is the product of in $Ho(Top)$. The same story works with $Top_\ast$, but now the product is the smash product of spaces.
The lack of colimits and limits is remedied in this context by considering homotopy limits and homotopy colimits. The idea here is that if you have a diagram in $Ho(M)$ then that's like a diagram in $M$ where we don't care about replacing objects up to weak equivalence. Given that choice not to care, the homotopy (co)limit is the object completing the diagram, i.e. satisfies a universal property which factors in the ability to replace objects in the diagram by weakly equivalent ones. Here is a good blog post on the subject.
The desire for nice categorical properties was one of the reasons people started to study the stable homotopy category. A great reference here is Margolis's book "Spectra and the Steenrod Algebra". In the introduction he defines the stable homotopy category in several steps and discusses which limits and colimits you are getting at each step. He mentions that even the category of CW complexes does not have good limit structures, and proceeds to move from CW to the Spanier-Whitehead category (which has stable homotopy classes of maps), and then go one farther to $\mathcal{S}$ (the coproduct completion of the subcategory $SW_f$ of finite complexes, a triangulated category). He discusses in chapter 3 that $\mathcal{S}$ has arbitrary coproducts and arbitrary weak colimits. In chapter 5, section 2, he mentions that it has arbitrary products and discusses weak limits.
The inclusion $\mathbb RP^2\to \mathbb CP^2$ has no kernel in the pointed homotopy category.
Proof:
Suppose $X\to \mathbb RP^2$ is such a kernel. Then maps $S^1\to X$ (in that category) correspond bijectively to maps $S^1\to\mathbb RP^2$ such that the composed map $S^1\to \mathbb CP^2$ is trivial. This means that $\pi_1(X)$ has order $2$.
It follows that there is a map $\mathbb RP^2\to X$ inducing an isomorphism of $\pi_1$. The composed map $\mathbb RP^2\to X\to \mathbb RP^2$ also induces an isomorphism of $\pi_1$, therefore also of $H^1$ with $\mathbb Z/2$ coefficients, therefore (using naturality of cup products) an iso of $H^2$ with $\mathbb Z/2$ coefficients.
But the map $X\to \mathbb RP^2$ must be zero on $H^2$ because the inclusion $\mathbb RP^2\to\mathbb CP^2$ is an isomorphism on $H^2$ and the composed map $X\to \mathbb RP^2 \to \mathbb CP^2$ is zero.
EDIT: Also, the degree two map $S^1\to S^1$ has no cokernel. Briefly, any such cokernel would have to be a retract (up to homotopy) of $\mathbb RP^2$, but that would make it either a point or $\mathbb RP^2$ and neither of these does the job.
A very concrete example of a cospan having no pullback in the homotopy category, which does not require any knowledge of cohomology or Moore spaces, can be found here. It's phrased in terms of the homotopy category of groupoids, but can easily be translated to speak about $K(\pi,1)$s in the homotopy category of spaces instead.