= Solution
Suppose a nontrivial finite <group> acts freely on $\mathbb R^n$ by <homeomorphisms>. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.
Let $M=\mathbb R^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-<manifold> with <universal cover> $\mathbb R^n$, which is contractible. The <group>-<cohomology> argument of Question 2 applies even if no cell structure has been chosen: <singular simplices> in $M$ lift to the cover, their lifts form free deck-<group> orbits, and the augmented <singular chains> of the contractible cover form an exact <free resolution>. Therefore
$$
H^i(M;\mathbb Z)\cong
\operatorname{Ext}_{\mathbb Z[C_p]}^i(\mathbb Z,\mathbb Z).
$$
We calculate the right-hand side explicitly. Write $R=\mathbb Z[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots+t^{p-1}$. The <periodic resolution of a finite cyclic group> is
$$
\cdots\xrightarrow{\,t-1\,}R\xrightarrow{\,N\,}R
\xrightarrow{\,t-1\,}R\xrightarrow{\varepsilon}\mathbb Z\longrightarrow0.
$$
Its exactness can be seen in coefficients. For $r=\sum_{j=0}^{p-1}a_jt^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker(t-1)=\mathbb ZN=\operatorname{im}N$. Also $Nr=\varepsilon(r)N$, so $\ker N=\ker\varepsilon$. If $\varepsilon(r)=0$, then
$$
r=\sum_{j=1}^{p-1}a_j(t^j-1)
=(t-1)\sum_{j=1}^{p-1}a_j(1+t+\cdots+t^{j-1}).
$$
Hence $\ker N=\ker\varepsilon=(t-1)R$, proving exactness everywhere.
Apply $\operatorname{Hom}_R(-,\mathbb Z)$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The <cochain complex> is therefore
$$
\mathbb Z\xrightarrow{0}\mathbb Z\xrightarrow{p}\mathbb Z
\xrightarrow{0}\mathbb Z\xrightarrow{p}\cdots.
$$
Its positive even-degree <cohomology> is
$$
H^{2j}(M;\mathbb Z)\cong\mathbb Z/p\mathbb Z
\quad(j\geq1),
$$
while its positive odd-degree <cohomology> is zero.
This contradicts the dimension of the quotient <manifold>. The noncompact form of <Poincaré duality for noncompact manifolds>, with the <orientation local system> when needed, is
$$
H^i(M;\mathbb Z)\cong
H_{n-i}^{\mathrm{BM}}(M;\mathcal O_M).
$$
Here <Borel-Moore homology> is the <homology (mathematics)> of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb Z)=0$ for $i>n$, whether or not $M$ is <orientable>. Taking any $2j>n$ contradicts the displayed <cyclic group> calculation. Thus
$$
\boxed{\text{a finite group acting freely on }\mathbb R^n
\text{ must be trivial}.}
$$
This proves <finite groups cannot act freely on Euclidean space> for general <homeomorphisms>; it does not assume that the action consists of affine maps or isometries.
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