= Solution
{c}
There are two <tree-level Feynman diagrams>, with intermediate <photon> and <gluon> respectively. Their <color charge> tensors are $\delta_{ji}\delta_{kl}$ and $\sum_AT^A_{ji}T^A_{kl}$. The pure <photon> factor is
$$
C_\gamma=\frac1{N_c^2}\sum_{i,j,k,l}|\delta_{ji}\delta_{kl}|^2=1.
$$
Their interference vanishes after the <color charge> sum because of <photon-gluon colour orthogonality>:
$$
C_{\gamma g}=\frac1{N_c^2}\sum_A\operatorname{tr}(T^A)\operatorname{tr}(T^A)=0.
$$
Thus there is no term proportional to $e^2g_s^2$, and with the $F_u$ of part (iii),
$$
\boxed{\overline{|\mathcal M_\gamma+\mathcal M_g|^2}
=\frac{2F_u}{s^2}\left[e^4Q_u^2Q_t^2+\frac29g_s^4\right]
=\frac{F_u}{s^2}\left[\frac{32}{81}e^4+\frac49g_s^4\right].}
$$
Here $Q_u=Q_t=2/3$. The absence of interference concerns the specified averaged and summed result; a fixed choice of external <color charge> need not have zero interference.
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