= Solution
For a based space $(X,x_0)$ and $k\geq1$, the <homotopy group> $\pi_k(X,x_0)$ is the set of based <homotopy> classes of maps $(I^k,\partial I^k)\to(X,x_0)$, equivalently based maps $S^k\to X$. Concatenating along one coordinate defines its group operation; reversing that coordinate gives the inverse. The group is abelian for $k\geq2$. The symbol $\pi_0(X)$ records path components and is generally only a pointed set. An <Eilenberg–MacLane space> $K(G,n)$ is a connected CW space with $\pi_n=G$ and all other positive <homotopy groups> zero; for $n>1$ the group $G$ must be abelian.
Projection onto coordinates gives
$$
\pi_k(X\times Y)\cong\pi_k(X)\times\pi_k(Y):
$$
a map to a product is exactly a pair of maps, and a based <homotopy> to the product is exactly a pair of based homotopies. These identifications respect concatenation. Thus $K(G,n)\times K(H,n)$ has exactly the <homotopy groups> required of $K(G\times H,n)$. Taking CW models and using the allowed uniqueness of Eilenberg–MacLane spaces gives
$$
\boxed{K(G,n)\times K(H,n)\simeq K(G\times H,n).}
$$
Let $p$ be prime. Take $S^\infty$ in complex coordinates, with $C_p$ acting by scalar multiplication by a primitive $p$th root of unity. This is a free action. Every map from a finite-dimensional sphere lands in a finite stage and becomes null-homotopic in a sufficiently higher sphere, so $S^\infty$ is weakly contractible; the <Whitehead theorem> for <CW complexes> makes it contractible. Therefore the infinite lens space $S^\infty/C_p$ has contractible <universal cover>, <fundamental group> $C_p$, and no higher <homotopy>: it is a model of $K(C_p,1)$.
Here is the <group cohomology> computation, including the ring structure. For the cyclic group generated by $g$, the standard free resolution alternates the maps $g-1$ and $N=1+g+\cdots+g^{p-1}$. Applying $\operatorname{Hom}_{\mathbb Z[C_p]}(-,\mathbb F_p)$ makes both maps zero, so each <cohomology> degree is one-dimensional. The period-two identity <chain map> represents a class $b$ in degree two, and cup multiplication by this periodicity class is an isomorphism from each degree to the degree two higher. For odd $p$, a nonzero degree-one class $a$ has $a^2=0$ by graded commutativity, while $b^j$ and $ab^j$ are nonzero by periodicity. For $p=2$ the model is <homotopy> equivalent to $\mathbb{RP}^\infty$, whose ring is the infinite version of that computed in question 2. Consequently the <mod-p cohomology ring of a cyclic group> is
$$
\boxed{H^*(K(C_p,1);\mathbb F_p)=
\begin{cases}\mathbb F_2[t],&p=2,\quad |t|=1,\\
\Lambda(a)\otimes\mathbb F_p[b],&p\text{ odd},\quad |a|=1,\ |b|=2.
\end{cases}}
$$
The standard results used are the contractible-cover description of a $K(G,1)$, its identification of singular <cohomology> with group <cohomology>, and the cyclic periodic resolution with cup-periodicity just stated.
The product equivalence and the assumed <Künneth theorem> over the <field> give
$$
\boxed{H^*(K(C_p\times C_p,1);\mathbb F_p)=
\begin{cases}\mathbb F_2[t_1,t_2],&p=2,\\
\Lambda(a_1,a_2)\otimes\mathbb F_p[b_1,b_2],&p\text{ odd}.
\end{cases}}
$$
The tensor product is graded: for odd $p$, $a_1a_2=-a_2a_1$ and each $a_i^2=0$. The dimensions in degree $d$ are $d+1$: for $p=2$ count $t_1^it_2^{d-i}$; for odd $p$ the generating series is $(1+t)^2/(1-t^2)^2=(1-t)^{-2}$.
If an $n$-dimensional <CW complex> has exactly one cell in dimension $n+1$, then $C^{n+1}(X;\mathbb F_p)=\mathbb F_p$. <Cellular cohomology> is the quotient of the kernel of its outgoing <coboundary> by the image of the incoming one, a quotient of a subspace of this one-dimensional vector space. Thus
$$
\boxed{H^{n+1}(X;\mathbb F_p)\text{ is either }0\text{ or }\mathbb F_p.}
$$
This <one-cell bound on next-degree cohomology> is valid even with infinitely many higher cells; the printed finite-per-dimension hypothesis is more than it needs.
Now a free finite-group action makes $S^n\to M=S^n/G$ a covering. For $n>1$ the sphere is simply connected, so $\pi_1(M)=G$, $\pi_i(M)=0$ for $1<i<n$, and $\pi_n(M)=\mathbb Z$. The covering also identifies every higher <homotopy group> with that of $S^n$. Use the <homotopy-group effect of attaching higher cells>: attaching $(i+1)$-cells, $i\geq2$, preserves <homotopy groups> below $i$ and quotients $\pi_i$ by the subgroup generated by the attaching classes and their fundamental-group translates. Attach a single $(n+1)$-cell along a generator of $\pi_n(M)$; its translates do not require additional cells since the original generator already generates the underlying group. This kills $\pi_n$ without changing $\pi_1$ or the lower groups.
Successively in each higher degree attach cells representing generators of the remaining <homotopy group>. In the union every higher <homotopy group> vanishes, since a map and its relevant <homotopy> have compact domain and are handled at a finite stage. The result is a $K(G,1)$ with exactly one $(n+1)$-cell. No finiteness assumption on the number of later cells is needed for the preceding <cochain> bound. If desired, one can retain finitely many per degree using the <Serre finiteness theorem for homotopy groups>: a simply connected <CW complex> of finite type has finitely generated <homotopy groups>. At each finite stage the <universal cover> is simply connected and finite type, since $G$ is finite; choosing finitely many generators then preserves finite type inductively. The first stage needs just the stated $\pi_n(S^n)=\mathbb Z$.
If $G=C_p\times C_p$, the resulting $K(G,1)$ would have $\dim H^{n+1}\leq1$ by its single cell, but its already computed <cohomology> has dimension $n+2>1$ in that degree. This contradiction proves
$$
\boxed{G\not\cong C_p\times C_p\qquad\text{for every prime }p.}
$$
This is the <free sphere actions exclude elementary abelian subgroups of rank two> obstruction.
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