= Solution
There are four relevant <tree-level Feynman diagrams>: $s$-channel <photon>, <Z boson> and <Higgs boson> exchange, and $t$-channel electron-neutrino exchange. The <electron> and <Positron> are incoming, and $W^-$ and $W^+$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.
\Image[/past-exam-of-the-gauge-field-theory-course-of-the-university-of-cambridge/2016/paper-1-w-pair-diagrams.png]
{title=The four tree-level diagrams for electron–positron annihilation into a charged W pair}
{description=Original schematics. Fermion arrows show particle-number flow, which opposes the momentum of an incoming positron. W lines are labelled by charge.}
Let $v=2m_W/g$. The <Higgs-electron coupling> is $m_e/v$, while the <Higgs boson coupling to W bosons> is $2m_W^2g_{\mu\nu}/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,
$$
\mathcal M_h=\frac{g^2m_e}{2(s-m_h^2)}\bar v(p_2)u(p_1)\,\epsilon_1^*\cdot\epsilon_2^*.
$$
A scalar propagator has no vector index, the <Yukawa interaction> supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.
For unpolarized incoming beams the <spin average> is $1/4$, not a sum without normalization. The <fermion spin sums> and <gamma-matrix trace> identities give
$$
\begin{aligned}
\frac14\sum_{r,t}|\bar v_t(p_2)u_r(p_1)|^2
&=\frac14\operatorname{tr}\left[(\not p_2-m_e)(\not p_1+m_e)\right]\\
&=p_1\cdot p_2-m_e^2
=\frac{s-4m_e^2}{2}.
\end{aligned}
$$
The terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.
In the <centre-of-momentum frame>, take $k_1=(E,k\widehat{\mathbf n})$ and $k_2=(E,-k\widehat{\mathbf n})$, where $E=\sqrt s/2$ and $k^2=E^2-m_W^2$. A pair of <longitudinal polarization vectors> is
$$
\epsilon_{1L}=\frac1{m_W}(k,E\widehat{\mathbf n}),\qquad
\epsilon_{2L}=\frac1{m_W}(k,-E\widehat{\mathbf n}).
$$
Each obeys $\epsilon_i\cdot k_i=0$ and $\epsilon_i^2=-1$. Their contraction is angle independent:
$$
\epsilon_{1L}\cdot\epsilon_{2L}=\frac{k^2+E^2}{m_W^2}=\frac{s}{2m_W^2}-1.
$$
Consequently
$$
\boxed{\overline{|\mathcal M_h|^2}_{LL}=\frac{g^4m_e^2(s-4m_e^2)}{8(s-m_h^2)^2}\left(\frac{s}{2m_W^2}-1\right)^2.}
$$
For the <two-body scattering cross section>, the outgoing-to-incoming momentum ratio is
$$
\frac{k}{p}=\frac{\beta_W}{\beta_e},\qquad
\beta_W=\sqrt{1-\frac{4m_W^2}{s}},\quad
\beta_e=\sqrt{1-\frac{4m_e^2}{s}}.
$$
The source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the <solid angle> gives $4\pi$:
$$
\begin{aligned}
\sigma_{h,LL}
&=\frac1{16\pi s}\frac{\beta_W}{\beta_e}\overline{|\mathcal M_h|^2}_{LL}\\
&=\boxed{\frac{g^4m_e^2\beta_e\beta_W}{128\pi(s-m_h^2)^2}\left(\frac{s}{2m_W^2}-1\right)^2},\qquad s\ge4m_W^2.
\end{aligned}
$$
There is no identical-particle factor: $W^+$ and $W^-$ are distinguishable. The threshold factor $\beta_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.
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