= Solution
The massless two-body <kinematics> gives $q_1\cdot q_2=M^2/2$. The <scalar> <scattering amplitude> therefore has constant magnitude $|\mathcal A_{aa}|=|\lambda_a|M^2/2$. Use $\int d\Omega=4\pi$ and the identical-particle factor from part d:
$$
\boxed{\Gamma_{aa}=\frac1{2!}\frac{4\pi}{64\pi^2M}\frac{|\lambda_a|^2M^4}{4}
=\frac{|\lambda_a|^2M^3}{128\pi}.}
$$
For the <photons>, the <two-photon vertex of a scalar field-strength interaction> following from $\Phi F_{\mu\nu}F^{\mu\nu}$ is, up to an irrelevant overall phase,
$$
\mathcal A_{\gamma\gamma}=\lambda_\gamma\left[(q_1\cdot q_2)(\epsilon_1\cdot\epsilon_2)
-(q_1\cdot\epsilon_2)(q_2\cdot\epsilon_1)\right].
$$
\b[Vertex qualification:] the source prints same-index <polarization> contractions in the second product, whereas the contractions must be crossed as above. The printed expression fails the <Ward identity>. For a concrete counterexample take $q_1=(E,0,0,E)$, $q_2=(E,0,0,-E)$ and $\epsilon_2=(1,1,0,-1)$. This is a transverse unit spacelike <photon polarization vector>, since $q_2\cdot\epsilon_2=0$ and $\epsilon_2^2=-1$. Replacing $\epsilon_1$ by $q_1$ in the printed expression leaves $4\lambda_\gamma E^3$ instead of zero. The corrected crossed expression vanishes under either such replacement.
Choose physical spatial transverse <photon polarizations> along $x$ and $y$ in the parent <rest frame>. Then both crossed products vanish; the two equal-polarization combinations have $|\epsilon_1\cdot\epsilon_2|^2=1$ and the two orthogonal combinations give zero. There is no initial-spin average for a spin-zero parent. Hence
$$
\sum_{\mathrm{pol}}|\mathcal A_{\gamma\gamma}|^2=2|\lambda_\gamma|^2\left(\frac{M^2}{2}\right)^2
=\frac{|\lambda_\gamma|^2M^4}{2},
\qquad
\boxed{\Gamma_{\gamma\gamma}=\frac{|\lambda_\gamma|^2M^3}{64\pi}.}
$$
The calculation in this spatial transverse gauge also yields the intended result from the misprinted expression, but only the corrected vertex makes its <gauge invariance> explicit. For equal-magnitude <couplings>, the <photon> <decay width> is twice the two-scalar <decay width> because of its two nonzero <polarization> combinations.
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