I do not understand something about the intersection product. I'm kind of new to this topic, so please consider that. I write down everything we discussed in a lecture.
We defined the intersection product of a connected and compact complex manifold $X$ of dimension $n$ to be the pairing $H^q(X,\mathbb{R}) \times H^p(X, \mathbb{R}) \to H^{p+q}(X,\mathbb{R})$ defined by $([\omega], [\varphi]) \mapsto [\omega \wedge \varphi]$, where the cohomology groups are identified with the corresponding de Rham cohomology group. In particular, if $p = 2n-q$, we can see the intersection product as the pairing $[\omega].[\varphi] = \int_X \omega \wedge \phi$. So far so good.
If $Z$ is an $m$-dimensional submanifold of $X$, we defined the fundamental class of $Z$ to be the linear form $[Z] \colon H^{2m}(X,\mathbb{R}) \to \mathbb{R}$ with $[\omega] \to \int_Z \omega$. Suppose $Y$ is another submanifold of $X$ with $\dim Y = n-m$.
First question is: What exactly is $[Z].[Y]$? I believe that we identify $H^{2m}(X,\mathbb{R})^\vee$ with $H^{2n-2m}(X, \mathbb{R})$ via Poincaré duality (the same for $Y$) and then take the intersection product defined above.
Second question: We have $[Z].[Y] = \#(Z \cap Y)$ if the intersection is finite. Why? Where can I find a proof?
To the next question: If we have $X = \mathbb{P}_n$, then we stated that $H^2(X, \mathbb{R})$ is generated by the fundamental class of a hyperplane $H$, i.e. by $h := [H]$, such that $h^n = [H_1]. \dots .[H_n] = 1$. Unfortunately, it is not stated what $H_1$, ..., $H_n$ are and why the above statement holds.
I would be glad if you help me, as I'm a bit slow on the uptake. You do not have to explain everything in detail, but I would really appreciate it if you show me the right way to think of these things.
Greetings and thanks in advance!
A good reference for all your questions is the book of Bott-Tu, Differential Forms in algebraic topology. I'll give here some idea of what the answers are, but for the details you really need to go to the book.
1) If you assume that $Z$ and $Y$ are trasversal, then $[Z][Y]$ is the class of $Z \cap Y$ which is a submanifold of $X$ by the transversality condition. It corresponds with what you do since the Poincaré dual of $Z \cap Y$ is the wedge product of the Poincaré duals of $Z$ and $Y$, so one can define it in that way more generally, although the transversality condition is important for all the other questions you pose.
2) Here you really need not only that the intersection is finite but also that is transversal. Then [Z].[Y] is the class of $\#(Z \cap Y)$ points and you are done.
3) The $H_i$ are general hyperplanes. $H^q({\mathbb P}_n)\cong {\mathbb R}$ for $q=2k$, $0\leq k \leq n$ and it vanishes for all other $q$: this can be easily shown inductively by a Mayer-Vietoris argument. It is clear that all Poincaré duals of the $H_i$ are cohomologous, since there is an automorphism of ${\mathbb P}_n$ homotopic to the identity mapping any chosen hyperplane to any other chosen hyperplane. Let $h$ be the Poincaré dual of any hyperplane. Since the wedge product of cohomology classes is Poincaré dual of the trasversal intersection of manifolds, the Poincaré dual of any linear space of codimension $k$, $[H_1] \cdots [H_k]$, is $h^{\wedge k}$. So $h^{\wedge n}$ is the Poincaré dual of a point, which is then by definition of Poincaré dual a generator of $H^{2n}({\mathbb P}_n)$ whose integral is $1$.
The main point above is that the Poincaré dual of transversal intersections is the wedge product. This requires localization principle, tubular neighbourhoods and Thom classes of normal bundle, so I suggest you look at the book for that.