= Solution
In a <manifold chart>, write a smooth <differential form> as $\eta=\sum_I a_I\,dx^{i_1}\wedge\cdots\wedge dx^{i_p}$, with $I$ increasing. Define its <exterior derivative> by
$$
d\eta=\sum_{I,j}\partial_j a_I\,dx^j\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_p}.
$$
To check that this definition is global, change coordinates to $y$. The <chain rule> expresses $dx^i$ as $\sum_j(\partial_jx^i)dy^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $dy^k\wedge dy^j$ is antisymmetric. The coordinate definition satisfies the graded <Leibniz rule>, so differentiating the transformed expression for $\eta$ leaves exactly $\sum_I da_I\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_p}$. This agrees with the original formula and proves independence of the <manifold chart>. The same cancellation of mixed derivatives gives $d^2=0$.
The <de Rham cohomology> is therefore
$$
H^p_{\mathrm{dR}}(M;\mathbb R)=
\frac{\ker(d:\Omega^p(M)\to\Omega^{p+1}(M))}
{\operatorname{im}(d:\Omega^{p-1}(M)\to\Omega^p(M))},
$$
the <vector space> of <closed differential forms> modulo <exact differential forms>.
Here is a <Green-theorem proof of exactness on a disc>. Identify the disc with a round disc centred at zero, and let $\eta=P\,dx+Q\,dy$ be a <closed differential form>. Set $f(x)=\int_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, <Green theorem> on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since $\partial_xQ-\partial_yP=0$. Hence
$$
f(x+h)-f(x)=\int_{[x,x+h]}\eta
=\int_0^1\eta_{x+sh}(h)\,ds.
$$
Taking $h=\varepsilon v$ and dividing by $\varepsilon$ yields $df_x(v)=\eta_x(v)$. The radial integral defining $f$ is smooth, so $\eta=df$ and
$$
\boxed{H^1_{\mathrm{dR}}(\Delta;\mathbb R)=0}.
$$
Degenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.
Integration requires an <orientation>: take $M$ to be an oriented compact $n$-dimensional <smooth manifold> without boundary. The <Generalized Stokes theorem> says $\int_Md\beta=0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an <exact differential form> leaves its integral unchanged. Thus
$$
[\eta]\longmapsto\int_M\eta
$$
is a well-defined <linear map> $H^n_{\mathrm{dR}}(M;\mathbb R)\to\mathbb R$. Compactness alone does not supply the orientation omitted from the printed integration statement.
For an $n$-dimensional <Lie group>, choose a nonzero $\lambda\in\Lambda^nT_e^*G$ and define the <invariant volume form on a Lie group> by
$$
\omega_g(v_1,\ldots,v_n)
=\lambda\bigl(d(L_{g^{-1}})_gv_1,\ldots,d(L_{g^{-1}})_gv_n\bigr).
$$
Each $dL_{g^{-1}}$ is an <isomorphism>, so $\omega$ is nowhere zero and defines an <orientation>. The composition identity $L_{(hg)^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^*\omega=\omega$. Moreover, every <left-invariant differential form> of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.
Under the supplied invariant-representative assumption, all <top-degree de Rham cohomology> classes are multiples of $[\omega]$. For compact $G$, orient it so $\omega$ is positive; then $\int_G\omega>0$, whereas the <Generalized Stokes theorem> makes every exact top form have integral zero. Therefore $[\omega]\ne0$ and
$$
\boxed{H^n_{\mathrm{dR}}(G;\mathbb R)\cong\mathbb R}
$$
under that assumption. For actual compact <Lie groups>, this argument applies when $G$ is <connected>. The supplied assumption fails for disconnected groups: on $S^1\times\mathbb Z/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the <top-degree de Rham cohomology of a compact Lie group> is $\mathbb R^c$, where $c$ is its number of <connected components>.
For $S^2$, use $U=S^2\setminus\{N\}$ and $V=S^2\setminus\{S\}$. <Stereographic projection> identifies each with the plane. The disc argument gives $\eta|_U=df_U$ and $\eta|_V=df_V$ for any <closed differential one-form> $\eta$. The overlap $U\cap V$ is <connected>, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global <smooth function> $f$ with $df=\eta$. Thus
$$
\boxed{H^1_{\mathrm{dR}}(S^2;\mathbb R)=0}.
$$
Finally, $S^3$ is a compact connected <Lie group>: the identification
$$
(a,b)\longmapsto\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix},
\qquad |a|^2+|b|^2=1,
$$
identifies it smoothly with the <special unitary group> $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives
$$
\boxed{H^3_{\mathrm{dR}}(S^3;\mathbb R)\cong\mathbb R}.
$$
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