computing $\pi_1(\mathbb{R}P^2 \vee\mathbb{R}P^2)$ and $\pi_1(\mathbb{R}P^2 \times \mathbb{R}P^2)$ could someone please check my solution? I'm trying to learn algebraic topology on my own.
First, I calculate the fundemantal group of $\mathbb{R}P^2$.
We know that $\mathbb{Z}/2 \rightarrow S^2\rightarrow \mathbb{R}P^2$ is a fibre sequence. Hence we have a long exact homotopy sequence $\dots\rightarrow \pi_2(\mathbb{R}P^2) \rightarrow \pi_1(\mathbb{Z}/2)\rightarrow \pi_1(S^2)\rightarrow \pi_1(\mathbb{R}P^2)\rightarrow \pi_0(\mathbb{Z}/2)\rightarrow \pi_0(S^2)\rightarrow\dots$.
This becomes $\dots\rightarrow \pi_2(\mathbb{R}P^2) \rightarrow 0\rightarrow0\rightarrow \pi_1(\mathbb{R}P^2)\rightarrow \pi_0(\mathbb{Z}/2)\rightarrow 0\rightarrow\dots$.
Hence $\pi_1(\mathbb{R}P^2)\cong \pi_0(\mathbb{Z}/2) \cong \mathbb{Z}/2$.
Since $\mathbb{R}P^2$ is path connected, it follows that $\pi_1(\mathbb{R}P^2 \times \mathbb{R}P^2)\cong \pi_1(\mathbb{R}P^2) \times \pi_1(\mathbb{R}P^2)\cong \mathbb{Z}/2 \times \mathbb{Z}/2.$
To calculate $\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2)$, I use Van Kampen's Theorem with $A=B=\mathbb{R}P^2$ and $A \cap B \simeq \ast$ (i.e. it's contractible). Since $\pi_1(A \cap B)$ is trivial, we get that $\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2)\cong \mathbb{Z}/2 \ast \mathbb{Z}/2.$
My questions: 1. have I made any mistakes?
2. Why is $\pi_i(\mathbb{Z}/2)=0$, for $i \ge 1$?
3. is there another method besides Van Kampen Theorem to calculate $\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2)$?
 A: First of all, let me point out that in the last section of the exact sequence, you're dealing with pointed sets and not groups. In particular, from $*\to \pi_1(\mathbb RP^2)\to \pi_0(\mathbb Z/2)\to *$ being an exact sequence of pointed sets, you can't conclude as easily that $\pi_1(\mathbb RP^2)\cong \mathbb Z/2$.
For instance, for any nontrivial group $G$, there is an exact sequence of pointed sets $*\to G\to \{0,1\}\to *$. So you have to put in a little more work if you only want to use that exact sequence (the usual road to computing this $\pi_1$ is through covering theory; but if you use additional data, I think you can use the exact sequence as well)
Your computations for $\times$ and $\vee$ from the computation of $\pi_1(\mathbb RP^2)$ are correct.
$\pi_i(\mathbb Z/2) = 0$ for $i>0$ because each point in $\mathbb Z/2$ is, well, a point, so it has trivial homotopy groups. You should try to prove that.
As pointed out by William in the comments, you could use covering theory to compute $\pi_1(\mathbb RP^2\vee \mathbb RP^2)$ although I wouldn't be surprised if that turned out to be a mess (just look at the result you're supposed to get). Van Kampen is the usual road for this computation (and seeing the result, it's reasonable to expect it to be the only sensible road)
