Q:

Homotopy group of the conformal group

I would like to know which are the first three homotopy groups of the conformal group SO(4,2):
$$
\pi_n(SO(4,2))=? \quad n=1,2,3
$$

A:

According to ncatlab, the maximal compact subgroup of (the connected component of) $SO(4,2)$ is $SO(4)\times SO(2)$. Any connected Lie group retracts onto its maximal compact subgroup, so your question is about $SO(4)\times SO(2)$. 
Since $SO(2)$ is just a circle, we have $\pi_n(SO(2))=\mathbb{Z},0,0,\ldots$.
$SO(4)$ is a semi-simple group, $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z_2}$, so it is easy to compute the first homotopy groups. 
Indeed because the second homotopy of any simple group is trivial, and the third is $\mathbb Z$, we have $\pi_2(SO(4))=0$ and $\pi_3(SO(4))=\mathbb{Z}^2$ (for $n>4$, $\pi_3(SO(n))=\mathbb{Z}$, since then $SO(n)$ is actually simple). 
The fundamental group of $SO(n)$, $n\geq 3$ is $\mathbb{Z}_2$, as can be seen in this specific case from the isomorphism above.
Thus we find finally (for the connected component)
$$
\pi_n(SO(4,2))=\mathbb{Z}_2\times \mathbb{Z},0,\mathbb{Z}^2,\mathbb{Z}_2^2,\mathbb{Z}_2^2,\mathbb{Z}_{12}^2,\ldots,
$$
or more generally for $n>1$
$$
\pi_n(SO(4,2))=\pi_n(SU(2))\times \pi_n(SU(2))=\pi_n(S^3)\times \pi_n(S^3),
$$
where the homotopy groups of spheres can be found on Wikipedia. Note that we do not know all the homotopy groups of $S^3$ in full generality, as noted by Mariano in the comments.
Edit: fixed the 3rd homotopy group and added the general relation to $S^3$.

