= Solution
The <tree-level Feynman diagrams> are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary <QCD> <Lagrangian density>. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal M$ as the product of the stated <Feynman rules>.
\Image[/past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2018-gluon-fusion-and-yukawa-loop.png]
{title=The three gluon-fusion tree diagrams and the Yukawa fermion-loop correction}
{description=The first three panels show ordered quark exchange and three-gluon exchange. The last panel shows a directed closed fermion loop between two scalar insertions. Crossings without dots are not vertices.}
Suppressing external colour indices, the <quark> graphs are
$$
\mathcal M_t^{\mu\nu}=-g^2\bar u(p_1)T^aT^b\gamma^\mu\frac{\not r_t}{t}\gamma^\nu v(p_2),
\qquad
\mathcal M_u^{\mu\nu}=-g^2\bar u(p_1)T^bT^a\gamma^\nu\frac{\not r_u}{u}\gamma^\mu v(p_2).
$$
The massless <Dirac equations> give $\bar u\not p_1=0$ and $\not p_2v=0$. Consequently
$$
\frac{\bar u\not k_1\not r_t\not k_2v}{t}=-\bar u\not k_2v=\bar u\not k_1v,
\qquad
\frac{\bar u\not k_2\not r_u\not k_1v}{u}=-\bar u\not k_1v.
$$
The last equality uses $\bar u\not qv=0$. The quark-exchange sum is therefore $-g^2\bar u\not k_1[T^a,T^b]v$.
For the <gluon> graph, use all incoming <momenta> $(k_1,k_2,-q)$ in the three-gluon tensor:
$$
V^{\mu\nu\rho}=(k_1-k_2)^\rho g^{\mu\nu}
+(k_2+q)^\mu g^{\nu\rho}-(q+k_1)^\nu g^{\rho\mu}.
$$
Its double contraction is $k_{1\mu}k_{2\nu}V^{\mu\nu\rho}=s(k_2-k_1)^\rho/2$. The vertex and <gluon> <propagator> give
$$
\mathcal M_s^{\mu\nu}=-\frac{ig^2f^{abc}}s\,
\bar u\gamma_\rho T^cv\,V^{\mu\nu\rho},
\qquad
k_{1\mu}k_{2\nu}\mathcal M_s^{\mu\nu}
=ig^2f^{abc}\bar u\not k_1T^cv.
$$
Combining the graphs yields
$$
\boxed{k_1^\mu k_2^\nu\mathcal M_{\mu\nu}
=-g^2\bar u(p_1)\not k_1
\big([T^a,T^b]-if^{abc}T^c\big)v(p_2).}
$$
Thus the requested <spinor> proportionality factor is $-g^2\bar u(p_1)\not k_1v(p_2)$ when the colour matrix is displayed separately; retaining it inside the <spinor> bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal M$, rather than $i\mathcal M$, the entire result gains one factor of $i$.
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