= Solution
{c}
Set $s=(p_1+p_2)^2=m_h^2$, so $p_1^2=p_2^2=0$ and $p_1\cdot p_2=s/2$. The physical <polarization vectors> obey $\epsilon_1\cdot p_1=\epsilon_2\cdot p_2=0$. Write $X_{\mu\nu}\simeq Y_{\mu\nu}$ when they agree after contraction with these <polarization vectors>; terms proportional to $p_{1,\mu}$ or $p_{2,\nu}$ may then be discarded.
The <gamma matrix trace identities> kill the terms with an odd number of <gamma matrices>. The remaining three two-<momentum> terms and the term with three factors of $m$ are
$$
\begin{aligned}
\frac{T_{\mu\nu}}{4m}={}&m^2g_{\mu\nu}
+l_\mu(l-p_2)_\nu+(l-p_2)_\mu l_\nu-g_{\mu\nu}l\cdot(l-p_2)\\
&+l_\mu(l+p_1)_\nu+(l+p_1)_\mu l_\nu-g_{\mu\nu}l\cdot(l+p_1)\\
&+g_{\mu\nu}(l-p_2)\cdot(l+p_1)-(l-p_2)_\mu(l+p_1)_\nu+(l+p_1)_\mu(l-p_2)_\nu.
\end{aligned}
$$
In the $g_{\mu\nu}$ coefficient the terms linear in $l\cdot p_i$ cancel, leaving $m^2-l^2-s/2$. Among the other terms, the surviving mixed products of $l$ and $p_i$ also cancel. Consequently
$$
T_{\mu\nu}\simeq4m\left[4l_\mu l_\nu+p_{2,\mu}p_{1,\nu}+(m^2-s/2-l^2)g_{\mu\nu}\right].
$$
In the second <Feynman diagram>, exchange the external labels and change the dummy <loop momentum> $l\mapsto-l$. Its projected numerator and denominator become identical to those of the first <Feynman diagram>. Hence
$$
\boxed{\mathcal M'_{\mu\nu}=8iN_cye^2Q^2m\int\frac{d^4l}{(2\pi)^4}
\frac{4l_\mu l_\nu+p_{2,\mu}p_{1,\nu}+(m^2-s/2-l^2)g_{\mu\nu}}{D_0D_1D_2}.}
$$
With the common factor $8m$ extracted, the primitive integer coefficients are
$$
\boxed{A=4,\qquad B=1,\qquad C=1,\qquad \text{prefactor}=8iN_c\frac{\lambda_t}{\sqrt2}e^2e_t^2m_t.}
$$
Equivalently, extracting only $m$ gives $(A,B,C)=(32,8,8)$ and prefactor $iN_c\lambda_te^2e_t^2m_t/\sqrt2$. A common rescaling of all three coefficients changes the prefactor inversely. The physical <scattering amplitude> is $\epsilon_1^\mu\epsilon_2^\nu\mathcal M'_{\mu\nu}$; the projected integrand is not an identity for an arbitrary uncontracted off-shell <tensor>.
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