what can you say about the degree of $f:\mathbb{C}P^n \to \mathbb{C}P^n$ Any thoughts on this problem:
If $M$ and $N$ are simply-connected, $n$-dimensional manifolds, then $H^n(M;\mathbb{Z}) \cong \mathbb{Z} \cong H^n(N;\mathbb{Z})$. A map $f:M \to N$ induces a map $f^*:H^n(N;\mathbb{Z}) \to H^n(M;\mathbb{Z})$, which is to say: $f^*:\mathbb{Z} \to \mathbb{Z}$. Any such map is given by multiplication by an integer $d$, and this integer is known as the degree of $f$, denoted by $deg(f)$. I have 2 questions:
(1) Let $f:\mathbb{C}P^2 \to \mathbb{C}P^2$. is it possible that $deg(f)=8$? is it possible that $deg(f)=9$? what can you say about the degree of $f$.
(2) what can you say about the degree of $f:\mathbb{C}P^n \to \mathbb{C}P^n$.
I know that we need to use the generators for each cohomology rings, but that is my issue I'm having trouble with this kind of questions. any help is appreciated. thanx in advance. 

As mentioned in the comments, any map from $\mathbb{C}P^n$ to itself must have degree a perfect $n$th power.  But do all such degrees actually arise?  Yes.
Consider the map $f_k:\mathbb{C}P^n\rightarrow \mathbb{C}P^n$ given by $f([z_0:...z_n]) = [z_0^k:z_1^k:...:z_n^k]$.
This is well defined:  $f([\lambda z_0:...:\lambda z_n]) = [\lambda^k z_0^k:...:\lambda^k z_n^k] = [z_0^k:...z_n^k]$.
I claim the map $f_k$ has degree $k^n$.  As mentioned in the comments, it is enough to show $f_k^\ast:H^2(\mathbb{C}P^n)\rightarrow H^2(\mathbb{C}P^n)$ is multiplication by $k$.  This is probably easier done in terms of homology.
Consider the natural inclusion $S^2\cong \mathbb{C}P^1\subseteq \mathbb{C}P^n$ given by mapping $[z_0:z_1]\in \mathbb{C}P^1$ to $[z_0:z_1:0...:0]\in\mathbb{C}P^n$.  It is well known that this inclusion induces an isomorphism on on $H_2$.  Further, the map $f_k$ preserves this $S^2$.
Thus, we get a commutative diagram \begin{array}
SS^2 & \stackrel{f_k}{\longrightarrow} & S^2 \\
\downarrow & & \downarrow \\
\mathbb{C}P^n & \stackrel{f_k}{\longrightarrow} & \mathbb{C}P^n
\end{array}
Since the maps from $S^2$ to $\mathbb{C}P^n$ are isomorphisms on $H_2$, we reduce the problem to computing $(f_k)_\ast:H_2(S^2)\rightarrow H_2(S^2)$.  But from the usual identification of $\mathbb{C}P^1$ with $S^2$, we see that $f_k$ is the map which preserves each line of lattitude, wrapping it around itself $k$ times.  Thus $(f_k)_\ast$ is multiplication by $k$.
