= Solution
{c}
Let $g$ and $g'$ be the two <gauge couplings>. In <unitary gauge>, $\Phi=(0,0,v+h)^T$ and the <weak hypercharge> term in the <gauge covariant derivative> vanishes. Using $(T^a)_{bc}=-i\epsilon_{abc}$, the kinetic term becomes
$$
\frac12(D_\mu\Phi)^T(D^\mu\Phi)
=\frac12(\partial h)^2+\frac12g^2(v+h)^2[W_\mu^1W^{1\mu}+W_\mu^2W^{2\mu}].
$$
Thus, in the basis $(W^1,W^2,W^3,B)$, the <gauge-boson mass matrix> and its <eigenvalues> are
$$
\boxed{M_V^2=g^2v^2\operatorname{diag}(1,1,0,0),\qquad
m_{W^\pm}=gv=\frac{gm}{2\sqrt\lambda},\quad m_{W^3}=m_B=0.}
$$
Here $W^\pm=(W^1\mp iW^2)/\sqrt2$. The two massless neutral <gauge bosons> may be rotated to any orthonormal basis; no neutral mass term selects a unique mixing angle.
Expanding the same kinetic term to first order in $h$ gives
$$
\mathcal L_{hVV}=g^2vh[(W^1)^2+(W^2)^2]=2g^2vhW^+_\mu W^{-\mu}.
$$
Functional differentiation gives the <Feynman rules>
$$
\boxed{hW^+_\mu W^-_\nu:\quad 2ig^2v\,g_{\mu\nu}
=\frac{2im_W^2}{v}g_{\mu\nu},}
$$
and, in the real basis, $hW^a_\mu W^b_\nu:2ig^2v\delta^{ab}g_{\mu\nu}$ for $a,b\in\{1,2\}$. All such three-point <Feynman vertices> involving $W^3$ or $B$ vanish. The factor two in the real basis comes from the two identical <gauge boson> fields; in the charged basis it is already explicit in $\mathcal L_{hVV}$.
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