= Solution
The change of field and parameters separates the <Lagrangian density> in <Euclidean space> into
$$
\begin{aligned}
\mathcal L_{\mathrm{ren}}&=\frac12(\partial\phi)^2+\frac12m^2\phi^2
+\frac{\widetilde\lambda_{2k}}{(2k)!}\phi^{2k},\\
\mathcal L_{\mathrm{ct}}&=\frac12\delta Z(\partial\phi)^2+\frac12\delta m^2\phi^2
+\frac{\delta\lambda_{2k}}{(2k)!}\phi^{2k}.
\end{aligned}
$$
The two-leg <counterterm> <interaction vertex> is $-(\delta Zp^2+\delta m^2)$, with the overall <momentum conservation> delta function understood. With $D_0(p)=1/(p^2+m^2)$, the explicit first-order contribution to the <propagator> is
$$
\boxed{\delta D_{\mathrm{ct}}(p)
=-\frac{\delta Zp^2+\delta m^2}{(p^2+m^2)^2}.}
$$
The minus sign follows either from expanding $e^{-S_{\mathrm{ct}}}$ or from expanding $(p^2+m^2+\delta Zp^2+\delta m^2)^{-1}$.
For $k=2$, the loop correction is $-\widetilde\lambda_4T_D(m)D_0(p)^2/2$. Its <ultraviolet pole> from <2/c> is
$$
\delta D_{\mathrm{loop}}\big|_{\mathrm{pole}}
=\frac{g_4m^2}{32\pi^2\epsilon}\,D_0(p)^2,
\qquad \widetilde\lambda_4=\mu^{2\epsilon}g_4.
$$
There is no external $p^2$ dependence in the amputated <tadpole diagram>, so no one-loop <wave-function renormalization> is required. The <minimal subtraction scheme> therefore gives
$$
\boxed{\delta Z^{(1)}=0,\qquad
\delta m^{2(1)}=\frac{g_4m^2}{32\pi^2\epsilon}.}
$$
Only the pole is subtracted: including $1-\gamma_E+\log4\pi$ would impose a different finite subtraction prescription. The defined $\delta m^2$ is the coefficient in $Zm_0^2=m^2+\delta m^2$, not automatically the difference $m_0^2-m^2$ beyond this order.
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