= Solution
{c}
Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common <colour factor> $\delta_{ji}$. Define $i\mathcal M$ by the product of <Feynman rules>. With massless external <Dirac spinors> the two graphs give
$$
\mathcal M_{\mu\nu}=-e^2Q_u^2\bar u(p_3)\left[\gamma_\nu\frac{\not q_s}{s}\gamma_\mu+\gamma_\mu\frac{\not q_u}{u}\gamma_\nu\right]u(p_1).
$$
The massless <Dirac equation> gives $\not p_1u(p_1)=0$ and $\bar u(p_3)\not p_3=0$. Thus
$$
\frac{\not q_s\not p_2}{s}u(p_1)=\frac{\not q_s(\not q_s-\not p_1)}s u(p_1)=u(p_1).
$$
By <four-momentum conservation>, $p_2=p_3-q_u$, so
$$
\bar u(p_3)\frac{\not p_2\not q_u}{u}=\bar u(p_3)\frac{(\not p_3-\not q_u)\not q_u}{u}=-\bar u(p_3).
$$
The two terms cancel:
$$
\boxed{p_2^\mu\mathcal M_{\mu\nu}=-e^2Q_u^2\bar u(p_3)(\gamma_\nu-\gamma_\nu)u(p_1)=0.}
$$
This is the <Ward identity> for the incoming <photon>; replacing its <photon polarization vector> by its <four-momentum> produces zero. The outgoing <photon> obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical <scattering amplitude> is invariant under $\varepsilon^\mu(p_2)\mapsto\varepsilon^\mu(p_2)+c p_2^\mu$.
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