= Solution
The <circle group> is $U(1)=\{z\in\mathbb C:|z|=1\}$ under multiplication. For continuous finite-dimensional complex representations, its irreducibles are the one-dimensional <characters>
$$
\boxed{\chi_m(z)=z^m,\qquad m\in\mathbb Z.}
$$
A <group representation> is a continuous homomorphism to the invertible linear maps; it is irreducible when it has no proper nonzero invariant subspace. <Compactness> permits an invariant Hermitian <inner product> by averaging. Since the resulting unitary operators commute for $U(1)$, simultaneous diagonalization reduces an irreducible to a <character>, and the continuous <characters> of the circle have the stated integer winding number.
The group $SU(2)$ consists of unitary two-by-two complex matrices of <determinant> one. Each has the unique form
$$
\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix},
\qquad |a|^2+|b|^2=1,
$$
identifying it homeomorphically with the unit 3-sphere in $\mathbb R^4$. This is the <Spin group> $\operatorname{Spin}(3)$. A <unitary matrix> is conjugate to $\operatorname{diag}(e^{i\theta},e^{-i\theta})$, and its conjugacy class is determined by $0\le\theta\le\pi$, or equivalently its <trace> $2\cos\theta$.
The irreducible <representations of SU2> are $V_n=\operatorname{Sym}^n(\mathbb C^2)$ for integers $n\ge0$, with dimension $n+1$. The symmetric power can be realized as homogeneous degree-$n$ <polynomials> in two variables, with the induced linear action. On the diagonal torus its <eigenvalues> are $e^{i(n-2j)\theta}$ for $j=0,\ldots,n$, so its <character>, the <trace> of the representing operator, is
$$
\boxed{\chi_n(\theta)=\sum_{j=0}^ne^{i(n-2j)\theta}
=\frac{\sin((n+1)\theta)}{\sin\theta}.}
$$
At the endpoints use the continuous values $\chi_n(0)=n+1$ and $\chi_n(\pi)=(-1)^n(n+1)$. The highest-weight classification says this list is complete.
For the final action, multiplication of block matrices gives $(AB)_1=A_1B_1$, so conjugation is a linear <group representation>. Decompose the underlying three-dimensional space as $V_1\oplus V_0$. Its endomorphism space is
$$
\operatorname{End}(V_1)\oplus\operatorname{Hom}(V_0,V_1)
\oplus\operatorname{Hom}(V_1,V_0)\oplus\operatorname{End}(V_0).
$$
The two off-diagonal blocks are each $V_1$, since the invariant alternating form identifies $V_1^*$ with $V_1$. The scalar part of $\operatorname{End}(V_1)$ is $V_0$ and its traceless part is $V_2$: the same alternating form identifies traceless endomorphisms with symmetric tensors of degree two. The lower scalar block is a second $V_0$. Thus
$$
\boxed{M_3(\mathbb C)\cong2V_0\oplus2V_1\oplus V_2.}
$$
The dimensions add to $2+4+3=9$, and its <character> is $2\chi_0+2\chi_1+\chi_2$.
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