= Solution
{c}
The <Slavnov-Taylor identity> implies transversality of the full perturbative <gluon self-energy>:
$$
k_\mu\left(\Pi_{3g}^{\mu\nu,ab}+\Pi_{4g}^{\mu\nu,ab}+\Pi_{\mathrm{gh}}^{\mu\nu,ab}+\Pi_q^{\mu\nu,ab}\right)=0.
$$
The <quark> loop is separately transverse. This follows without evaluating its <Feynman integral>: contract its vector <Feynman vertex> with $k_\mu$ and use $\not k=(\not l+\not k-m)-(\not l-m)$ between its two <Dirac propagators>. One obtains a difference of two loop integrals that cancel after a shift of $l$ in <dimensional regularization>. The <colour factor> is proportional to $\operatorname{tr}(T^aT^b)$ and does not obstruct this argument.
The four-gluon <tadpole diagram> vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the <ghost loop> must cancel. Contracting the preceding <ghost-loop pole of the gluon self-energy> gives
$$
\boxed{k_\mu\Pi_{3g}^{\mu\nu,ab}\big|_{\mathrm{UV}}=-k_\mu\Pi_{\mathrm{gh}}^{\mu\nu,ab}\big|_{\mathrm{UV}}=-\frac{g^2C_A\delta^{ab}}{32\pi^2\epsilon}k^2k^\nu.}
$$
For <QCD>, set $C_A=3$. No evaluation of the remaining three-gluon <Feynman diagram> is required. Its uncontracted transverse part is not fixed by this identity.
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