= Solution
{c}
For <massless quark Compton scattering>, the <color average> contributes one, since $(1/N_c)\sum_{i,j}|\delta_{ji}|^2=1$. The <spin average> and incoming <photon> <polarization> average supply $1/(2\cdot2)=1/4$. The <Ward identities> permit use of $-g_{\mu\nu}$ for each <photon polarization completeness relation>, and the massless <fermion spin sum> is $\sum_ru_r(p)\bar u_r(p)=\not p$.
Let
$$
A_{\mu\nu}=\gamma_\nu\frac{\not q_s}s\gamma_\mu+\gamma_\mu\frac{\not q_u}u\gamma_\nu.
$$
Reversing the <gamma matrices> order gives its Dirac-conjugate matrix $\bar A^{\mu\nu}$. The squared <scattering amplitude> is
$$
\langle|\mathcal M|^2\rangle=\frac{e^4Q_u^4}{4}\operatorname{tr}(\not p_3 A_{\mu\nu}\not p_1\bar A^{\mu\nu}).
$$
For clarity evaluate the four traces before dividing by their denominators. The <Mandelstam variables> obey $s+t+u=0$, with $t=(p_1-p_3)^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the <gamma matrix trace identities> imply
$$
\begin{aligned}
T_{ss}&=\operatorname{tr}(\not p_3\gamma_\nu\not q_s\gamma_\mu\not p_1\gamma^\mu\not q_s\gamma^\nu)\\
&=-2\operatorname{tr}(\not p_3\gamma_\nu\not q_s\not p_1\not q_s\gamma^\nu)\\
&=-2s\operatorname{tr}(\not p_3\gamma_\nu\not p_2\gamma^\nu)
=4s\operatorname{tr}(\not p_3\not p_2)=-8su.
\end{aligned}
$$
Here $\not q_s\not p_1\not q_s=2(q_s\cdot p_1)\not q_s-q_s^2\not p_1=s\not p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $\not q_u\not p_1\not q_u=-u\not p_4$ gives
$$
T_{uu}=\operatorname{tr}(\not p_3\gamma_\mu\not q_u\gamma_\nu\not p_1\gamma^\nu\not q_u\gamma^\mu)=-16u(p_3\cdot p_4)=-8su.
$$
One interference trace reduces to
$$
\begin{aligned}
T_{su}&=\operatorname{tr}(\not p_3\gamma_\nu\not q_s\gamma_\mu\not p_1\gamma^\nu\not q_u\gamma^\mu)\\
&=-2\operatorname{tr}(\not p_3\gamma_\nu\not q_s\not q_u\gamma^\nu\not p_1)
=-32(q_s\cdot q_u)(p_3\cdot p_1)=0,
\end{aligned}
$$
since $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. Therefore
$$
\boxed{\langle|\mathcal M|^2\rangle=-2e^4Q_u^4\left(\frac us+\frac su\right)=-\frac{32e^4}{81}\frac{s^2+u^2}{su},\qquad C=\frac{32e^4}{81}.}
$$
Equivalently $C=512\pi^2\alpha_{\mathrm{em}}^2/81$ in terms of the <fine-structure constant>. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a <collinear divergence>; it is not a negative squared <scattering amplitude>.
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