= Solution
First derive the needed <gamma-matrix traces> directly from the <Clifford algebra>. The <cyclic property of the trace> gives
$$
2\operatorname{tr}(\gamma^\mu\gamma^\nu)
=\operatorname{tr}\{\gamma^\mu,\gamma^\nu\}
=2\eta^{\mu\nu}\operatorname{tr}I_4,
$$
so $\operatorname{tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}$. For $T=\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)$, move $\gamma^\mu$ successively through the other three matrices and cycle the final term back to the front:
$$
2T=2\eta^{\mu\nu}\operatorname{tr}(\gamma^\rho\gamma^\sigma)
-2\eta^{\mu\rho}\operatorname{tr}(\gamma^\nu\gamma^\sigma)
+2\eta^{\mu\sigma}\operatorname{tr}(\gamma^\nu\gamma^\rho).
$$
Therefore
$$
\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)
=4(\eta^{\mu\nu}\eta^{\rho\sigma}-\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho}).
$$
These are all the trace identities required.
For massless external <Dirac spinors>, the <fermion spin sums> are $\sum_su_s(p)\bar u_s(p)=\not p$ and $\sum_rv_r(q)\bar v_r(q)=\not q$. With the initial <spin average>, part (iii) becomes a contraction of
$$
L^{\mu\nu}=4(q^\mu p^\nu+q^\nu p^\mu-\eta^{\mu\nu}p\cdot q),\qquad
H_{\mu\nu}=4(p'_\mu q'_\nu+p'_\nu q'_\mu-\eta_{\mu\nu}p'\cdot q').
$$
Their contraction is
$$
L^{\mu\nu}H_{\mu\nu}
=32\bigl[(q\cdot p')(p\cdot q')+(q\cdot q')(p\cdot p')\bigr].
$$
The massless <Mandelstam variables> give $p\cdot p'=q\cdot q'=-t/2$ and $p\cdot q'=q\cdot p'=-u/2$, hence
$$
\overline{|\mathcal M|^2}=\frac{2e^4}{s^2}(t^2+u^2).
$$
In the <center of mass> frame each particle has energy $\sqrt s/2$, and
$$
t=-\frac s2(1-\cos\theta),\qquad u=-\frac s2(1+\cos\theta).
$$
Substitution gives the concise angular answer:
$$
\boxed{\overline{|\mathcal M|^2}=e^4(1+\cos^2\theta).}
$$
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