Which groups act freely on $S^n$?
Here are some thoughts. If a finite group $G$ acts on $S^n$ freely, then it's automatically a covering space action as $S^n$ is Hausdorff. Thus, $\pi_1(S^n/G) \cong G$. Moreover, $\pi_i(S^n/G) \cong \pi_i(S^n) \cong 0$ for all $i < n$. $\pi_n(S^n/G) \cong \pi_n(S^n)$ which is infinite cyclic. Pick a map $S^n \to S^n/G$ representing the generator, and glue a $D^{n+1}$ along it to kill $\pi_n$. Kill $\pi_k$ for all $k > n$ similarly by gluing cells along maps representing generators of $\pi_k$ for each dimension $k$, to infinity if necessary.
One then obtains a $K(G, 1)$ from attaching just one $(n+1)$-cell to $S^n/G$. Thus, the group cohomology $H^{n+1}(G; \Bbb Z/p)$ is either $\Bbb Z/p$ or $0$ for any prime $p$ cellular cohomology.
As a corollary, for example, $\Bbb Z/2 \times \Bbb Z/2$ cannot act on any $S^n$ freely for any $n$, as by Kunneth formula
$$H^{n+1}(\Bbb Z/2 \times \Bbb Z/2; \Bbb Z/2) \cong \bigoplus_{i + j = n+1} H^i(\Bbb Z/2; \Bbb Z/2) \otimes H^j(\Bbb Z/2; \Bbb Z/2)$$
which is simply isomorphic to $(\Bbb Z/2)^{n+1}$, violating the previous obstruction. Similarly, $\Bbb Z/p \times \Bbb Z/p$ does not act on $S^n$ for any prime $p$. However, similar technique doesn't work for infinite groups. Also, can any obstruction be placed over higher group cohomologies?
You have already given answers yourself. I only have a two comments, with a recent reference, which you might have already seen.
If a finite group $G$ acts freely on a sphere then we know that all abelian subgroups of $G$ are cyclic, i.e., that $G$ has periodic cohomology, and that all elements of order $2$ are central. In particular, $G$ has at most one element of order $2$. For even-dimensional spheres there is only $C_2$, see also here.
For infinite groups, free actions of discrete groups have been studied a lot. A free action of a discrete group $G$ on an $n$-homotopy sphere $\Sigma(n)$ induces an action on $H^n(\Sigma(n),\mathbb{Z})$, i.e., an homomorphism $G\rightarrow Aut(H^n(\Sigma(n),\mathbb{Z}))$. For $G$ finite, and $n$ odd, this action is trivial. If the group $G$ is infinite there are more possibilities for the induced action of $G$, which makes it more difficult to characterise these induced actions. For a summary of some results and a certain classification see the recent preprint on Free and properly discontinuous actions of groups on homotopy $2n$-spheres.
This has almost but not quite been stated a few times, so to clear the air: the answer is known for finite groups, it is due to Madsen, Thomas, and Wall, and it says that a finite group $G$ acts freely on some sphere if and only if
- all of the abelian subgroups of $G$ are cyclic; equivalently, the cohomology is periodic; equivalently, $\mathbb{Z}_p \times \mathbb{Z}_p$ does not occur as a subgroup for any prime $p$; and
- every element of order $2$ is central.
The necessity of the first condition is due to Smith and the necessity of the second condition is due to Milnor. This is taken from the introduction to Alejandro Adem's Constructing and Deconstructing Group Actions.