= Solution
The <Adjoint double cover from SU(2) to SO(3)> is obtained by writing a traceless Hermitian matrix as $X=x_i\sigma_i$, where the $\sigma_i$ are the <Pauli matrices>, and defining $AXA^\dagger=(R(A)x)_i\sigma_i$. Explicitly,
$$
R_{ij}(A)=\frac12\operatorname{tr}(\sigma_iA\sigma_jA^\dagger).
$$
Conjugation preserves $\tfrac12\operatorname{tr}X^2=|x|^2$, so $R(A)$ is orthogonal; connectedness of <SU(2)> gives $\det R(A)=1$. Composition of conjugations makes $R$ a <group homomorphism>. A matrix in its kernel commutes with every <Pauli matrix>, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp(-i\theta\widehat n\cdot\sigma/2)$ induces the rotation through angle $\theta$ about $\widehat n$. Thus
$$
\boxed{SO(3)\cong SU(2)/\{\pm I\}.}
$$
Since <SU(2)> is the simply connected three-sphere, this is the universal double cover; the two <Lie algebras> are isomorphic, although the <groups> are not.
Define the <conjugate fundamental spinor of SU(2)> by $\bar\eta_\alpha=(\eta^\alpha)^*$. Regard $\bar\eta$ as a row, so unitarity gives
$$
\boxed{\bar\eta'_\alpha=\bar\eta_\beta(A^\dagger)^\beta{}_\alpha=\bar\eta_\beta(A^{-1})^\beta{}_\alpha.}
$$
Consequently $\bar\eta_\alpha\xi^\alpha$ is an invariant Hermitian pairing. The basic <invariant tensors> are the <Kronecker delta> $\delta^\alpha{}_\beta$ and the alternating tensors $\epsilon_{\alpha\beta}$ and $\epsilon^{\alpha\beta}$. Choose $\epsilon_{12}=1$, $\epsilon^{12}=-1$, so that $\epsilon^{\alpha\gamma}\epsilon_{\gamma\beta}=\delta^\alpha{}_\beta$. The unit-determinant identity $A^T\epsilon A=\epsilon$ makes $[\eta,\xi]=\epsilon_{\alpha\beta}\eta^\alpha\xi^\beta$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular $\widetilde\eta^\alpha=\epsilon^{\alpha\beta}\bar\eta_\beta$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta\mapsto\widetilde\eta$ squares to $-I$, so this is a <pseudoreal representation>, rather than a real structure.
For <Representation theory of SU2>, take $V_n=\operatorname{Sym}^n\mathbb C^2$, with $A$ acting on every index of a <symmetric tensor>. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the <homogeneous polynomial representation of SU2> has basis $z_1^{n-r}z_2^r$, $0\leq r\leq n$, with
$$
H=z_1\partial_{z_1}-z_2\partial_{z_2},\qquad E_+=z_1\partial_{z_2},\qquad E_-=z_2\partial_{z_1}.
$$
The basis weights are $n,n-2,\ldots,-n$, and the <raising operator> and <lowering operator> connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an <irreducible representation>, of <spin> $j=n/2$. The <classification of finite-dimensional representations of SU2> gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1)^n$, so <descent of an SU(2) representation to SO(3)> gives
$$
\boxed{\dim V_n=n+1,\qquad V_n\text{ descends to }SO(3)\iff n\text{ is even}.}
$$
For the <tensor contractions in the SU2 Clebsch-Gordan decomposition>, contract $r$ indices of $S$ with $r$ indices of $T$ using $\epsilon$, and symmetrize all the remaining free indices:
$$
U_r^{\alpha_1\ldots\alpha_{m+n-2r}}=
\operatorname{Sym}_{\alpha_1,\ldots,\alpha_{m+n-2r}}
\left(\epsilon_{\beta_1\gamma_1}\cdots\epsilon_{\beta_r\gamma_r}
S^{\alpha_1\ldots\alpha_{m-r}\beta_1\ldots\beta_r}
T^{\alpha_{m-r+1}\ldots\alpha_{m+n-2r}\gamma_1\ldots\gamma_r}\right),
\qquad 0\leq r\leq\min(m,n).
$$
Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The <highest-weight vectors>
$$
F_r=(z_1w_2-z_2w_1)^r z_1^{m-r}w_1^{n-r}
$$
are killed by the total <raising operator> $z_1\partial_{z_2}+w_1\partial_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The <unitary representation> on this tensor product makes invariant complements available, and the dimension check
$$
\sum_{r=0}^{\min(m,n)}(m+n-2r+1)=(m+1)(n+1)
$$
exhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the <Clebsch-Gordan decomposition for SU2> is
$$
\boxed{V_m\otimes V_n\cong\bigoplus_{r=0}^{\min(m,n)}V_{m+n-2r}.}
$$
For the <decomposition of three SU(2) spinors>, write $Q^{\alpha\beta\gamma}=\eta_1^\alpha\eta_2^\beta\eta_3^\gamma$ and $S^{\alpha\beta\gamma}=Q^{(\alpha\beta\gamma)}$. The contractions
$$
\begin{aligned}
d_1^\alpha&=\epsilon_{\beta\gamma}Q^{\alpha\beta\gamma}=\eta_1^\alpha[\eta_2,\eta_3],\\
d_2^\alpha&=\epsilon_{\beta\gamma}Q^{\beta\alpha\gamma}=\eta_2^\alpha[\eta_1,\eta_3],\\
d_3^\alpha&=\epsilon_{\beta\gamma}Q^{\beta\gamma\alpha}=\eta_3^\alpha[\eta_1,\eta_2]
\end{aligned}
$$
transform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, is
$$
Q^{\alpha\beta\gamma}=S^{\alpha\beta\gamma}
+\epsilon^{\alpha\beta}\frac{2d_1^\gamma-d_2^\gamma}{3}
-\epsilon^{\alpha\gamma}\frac{d_1^\beta+d_2^\beta}{3}.
$$
Symmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus
$$
\boxed{V_1^{\otimes3}=V_3\oplus V_1\oplus V_1,
\qquad \tfrac12\otimes\tfrac12\otimes\tfrac12=\tfrac32\oplus\tfrac12\oplus\tfrac12.}
$$
In the <quark model>, three spin-$1/2$ <quarks> therefore have total quark <spin> $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state <baryons> with orbital angular momentum $L=0$, total $J=S$: the <nucleon> has $J=1/2$ and the <Delta baryon> has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric <three-quark colour singlet>, the spatial wavefunction and <flavor symmetry> must also satisfy the <Pauli constraint on three-quark flavour multiplets>; orbital excitations can give other total angular momenta.
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