= Solution
Because the bilinear form from part (ii) is invariant, $\rho(SU(2))\subseteq O(3)$. The map
$$
U\longmapsto\det\rho(U)\in\{1,-1\}
$$
is continuous. Since $SU(2)$ is connected and $\det\rho(I_2)=1$, it is identically one. Hence
$$
\rho(SU(2))\subseteq SO(3).
$$
To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation with
$$
A_3=\begin{pmatrix}i&0\\0&-i\end{pmatrix}
$$
forces $U$ to be diagonal. Commutation with
$$
A_2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
$$
then forces its two diagonal entries to agree. Thus $U=zI_2$. Since $U\in SU(2)$,
$$
1=\det U=z^2,
$$
and therefore
$$
\boxed{\ker\rho=\{I_2,-I_2\}.}
$$
It remains to prove surjectivity. The <derived representation> at the identity is
$$
d\rho_{I_2}(X)(A)=[X,A].
$$
If this vanishes for every $A$ in the <SU(2) Lie algebra>, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d\rho_{I_2}$ is injective. Its domain and codomain are both three-dimensional real <Lie algebras>, so it is an isomorphism
$$
\mathfrak{su}(2)\longrightarrow\mathfrak{so}(3).
$$
The <inverse function theorem> now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The <special orthogonal group> $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, so
$$
\boxed{\rho:SU(2)\twoheadrightarrow SO(3).}
$$
Together with the kernel calculation, this is the <Adjoint double cover from SU(2) to SO(3)>.
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