= Solution
{c}
For <fundamental-doublet breaking of SU(2)>, assume $\lambda>0$ and $v>0$. The minima of the <scalar potential> satisfy $\phi^\dagger\phi=v^2/2$, giving a three-sphere <vacuum manifold>. A chosen <vacuum expectation value> is $\phi_0=(0,v)^T/\sqrt2$. No nonzero generator $a_a\tau^a$ annihilates $\phi_0$: its squared norm is $v^2(a_1^2+a_2^2+a_3^2)/8$. At the group level, an $SU(2)$ matrix fixing the nonzero vector $\phi_0$ must be the identity. Thus all three generators of the <SU(2) gauge theory> are broken in the usual perturbative <Higgs mechanism> description.
Make the radial and angular fields explicit:
$$
\phi(x)=\exp\left(\frac{i\sigma^a\pi^a(x)}v\right)\frac1{\sqrt2}\begin{pmatrix}0\\v+h(x)\end{pmatrix}.
$$
A local <gauge transformation> removes the three angular fields $\pi^a$. In <unitary gauge>, the <Pauli matrices> identity $\{\tau^a,\tau^b\}=\delta^{ab}\mathbf1/2$ gives
$$
(D_\mu\phi)^\dagger D^\mu\phi=\frac12(\partial_\mu h)(\partial^\mu h)+\frac{g^2}{8}(v+h)^2A_\mu^aA^{a\mu}.
$$
The mixed derivative term cancels between a term and its complex conjugate. The <scalar potential> becomes
$$
V=\lambda\left(vh+\frac{h^2}{2}\right)^2=\lambda v^2h^2+\lambda vh^3+\frac\lambda4h^4.
$$
Consequently the complete physical <Lagrangian density> is
$$
\begin{aligned}
\mathcal L={}&-\frac14F_{\mu\nu}^aF^{a\mu\nu}+\frac12(\partial h)^2
+\frac12m_A^2A_\mu^aA^{a\mu}-\frac12m_h^2h^2\\
&+\frac{g^2v}{4}hA_\mu^aA^{a\mu}+\frac{g^2}{8}h^2A_\mu^aA^{a\mu}-\lambda vh^3-\frac\lambda4h^4,\\
F_{\mu\nu}^a={}&\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+g\epsilon^{abc}A_\mu^bA_\nu^c.
\end{aligned}
$$
Reading the quadratic terms with their canonical factors $1/2$ gives
$$
\boxed{m_{A^1}=m_{A^2}=m_{A^3}=\frac{gv}{2},\qquad m_h=\sqrt{2\lambda}\,v.}
$$
All physical scalar-vector and scalar self-interactions are displayed in the second line of $\mathcal L$. In particular the new cubic couplings proportional to $v$ are $hA^aA^a$ and $h^3$. Their <Feynman vertices>, together with the quartic ones, are
$$
\boxed{hA_\mu^aA_\nu^b:\ i\frac{g^2v}{2}\delta^{ab}g_{\mu\nu},\quad hhA_\mu^aA_\nu^b:\ i\frac{g^2}{2}\delta^{ab}g_{\mu\nu},\quad hhh:\ -6i\lambda v,\quad hhhh:\ -6i\lambda.}
$$
The three-vector and four-vector interactions are already contained in the non-Abelian <gauge field strength> term. There are no further physical fields or interactions in this specified model. Before fixing <unitary gauge>, the quadratic angular term is $\tfrac12(\partial_\mu\pi^a-m_AA_\mu^a)^2$. It exhibits three zero-curvature angular directions of the <scalar potential>, but also their mixing with the <gauge fields>. They supply the third <polarization> of each massive <gauge boson>, rather than massless physical <particles>.
The degrees of freedom count is $3\cdot2+4=10$ before breaking and $3\cdot3+1=10$ afterwards. If instead the symmetry were global, as in the decoupled $g=0$ scalar theory, the three angular fields would be physical massless <Goldstone bosons>. This is consistent with <Goldstone's theorem>. The language of a broken local symmetry refers to this gauge-fixed expansion; a gauge-dependent <vacuum expectation value> is not itself a gauge-invariant observable.
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