= Solution
With a single quartic <interaction vertex>, the two unused line ends close into one <tadpole diagram>. With a single sextic <interaction vertex>, the four unused ends pair into two <tadpole diagrams>. These are the only one-vertex <one-particle irreducible> <Feynman diagrams> for the <two-point function>.
Define the Minkowski <Feynman integrals>
$$
D(k)=\frac{i}{k^2-m^2+i0},\qquad
T_M=\int\frac{d^3k}{(2\pi)^3}D(k).
$$
The amputated <two-point function> contributions are
$$
\boxed{C_4(-ig_1)T_M,\qquad C_6(-ig_2)T_M^2,}
$$
where $C_4,C_6$ are the reciprocals of the corresponding <Feynman-diagram symmetry factors>. Both are independent of external <momentum>, and hence affect the <mass counterterm> rather than <wave-function renormalization>. No numerical <Feynman-diagram symmetry factors> are needed here.
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