For a topological space $X$, define $LX$ to be the set of continuous maps $S^1 \rightarrow X$ with the compact-open topology. Henceforth let $X = \Bbb{CP}^\infty \times \Bbb{RP}^\infty$, with basepoint $([1:0:\dots], [1:0:\dots])$. Inspired by this comment, I tried to calculate $\pi_1(LX)$. During the calculation I found that $\pi_2(LX) = \Bbb Z$ or $\pi_2(LX) = 0$; in the former case, $\pi_1(LX)$ is finite, and in the latter case, it's a semidirect product of $\Bbb Z$ and $\Bbb Z/2\Bbb Z$. So I'd like to know which of these $\pi_2$ is.
Is there any way to calculate whether or not $\pi_2(LX)$ is nontrivial?
$\mathbb{CP}^{\infty} \times \mathbb{RP}^{\infty}$, being a product of Eilenberg-MacLane spaces $B^2 \mathbb{Z} \times B \mathbb{Z}_2$, is in particular a loop space and hence is homotopy equivalent to a topological group $G$ (in fact a topological abelian group). As I mentioned in the other thread, for any topological group $G$ the free loop space splits as a product (of spaces)
$$LG \cong G \times \Omega G$$
(sketch: fix a point in $S^1$, and evaluate a map $S^1 \to G$ on it; this gives you a point in $G$ and a based loop at that point. But on a topological group $G$ you can canonically translate this based loop to a based loop at the identity), and so we can compute its first two homotopy groups to be
$$\pi_1(LG) \cong \pi_1(G) \times \pi_2(G), \pi_2(LG) \cong \pi_2(G) \times \pi_3(G).$$
In particular, $\pi_2(LG) \cong \mathbb{Z}$ and $\pi_1(LG) \cong \mathbb{Z}_2 \times \mathbb{Z}$ is a direct product, not a semidirect product.
In general, suppose $X$ is a (pointed) space and we'd like to compute the homotopy groups of $LX$ (based at the constant loop at the basepoint of $X$). There is a nice trick we can play here which I didn't mention in the previous thread because I hadn't figured it out yet: it turns out that the operation of taking based and free loops commute, so that
$$\Omega LX \cong L \Omega X$$
so now we can use the fact that $\Omega X$, being a loop space, is homotopy equivalent to a topological group and hence that its free loop space splits, at least up to homotopy, as a product (of spaces)
$$\Omega LX \cong L \Omega X \cong \Omega X \times \Omega^2 X.$$
Applying $\pi_i, i \ge 1$, we conclude that
$$\pi_{i+1}(LX) \cong \pi_{i+1}(X) \times \pi_{i+2}(X), i \ge 1$$
So that takes care of $\pi_2$ and higher. For $\pi_1$ I suspect, but am not yet sure how to prove, that it is the semidirect product
$$\pi_1(LX) \cong \pi_2(X) \rtimes \pi_1(X)$$
(the action of $\pi_1$ on $\pi_2$ is trivial in the example you give), and for $\pi_0$ we can say the following: $\pi_0(LX)$ breaks up based on the connected component in $\pi_0(X)$ that a given loop is based in, and in each connected component the free homotopy classes of loops are identified with the set of conjugacy classes in the fundamental group $\pi_1(X, x)$ based at any $x$ in the component.