= Solution
Take Euclidean <gamma matrices> with $\{\gamma_\mu,\gamma_\nu\}=2\delta_{\mu\nu}$, and independent <Grassmann fields> $\psi,\bar\psi$. The minimally coupled <Dirac action> is
$$
\boxed{S_\psi=\int d^4x\,\bar\psi^a(\gamma_\mu D_\mu+m)_{ac}\psi^c.}
$$
Its mass term is the <Dirac mass term>, and the interaction is $-ig\bar\psi^a\gamma_\mu A_\mu^b(T^b_{\mathrm{ad}})_{ac}\psi^c$. With Fourier convention $e^{ip\cdot x}$, the free kinetic operator is $i\not p+m$. The <Euclidean adjoint Dirac Feynman rules> are therefore
$$
\boxed{S^{ac}(p)=\delta^{ac}\frac{-i\not p+m}{p^2+m^2},\qquad V_\mu^{b;ac}=+ig\gamma_\mu(T^b_{\mathrm{ad}})_{ac}=gf^{bac}\gamma_\mu.}
$$
The vertex sign follows by expanding $e^{-S_\psi}$, not $e^{iS_\psi}$. <Momentum conservation> supplies the usual <Dirac delta function> at each vertex. There is only a one-gluon fermion vertex, with no two-gluon seagull because the fermionic <covariant derivative> is first order. A closed fermion loop has the usual <fermionic sign> $-1$. The identity $(i\not p+m)(-i\not p+m)=(p^2+m^2)I$ checks the <Dirac propagator>.
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