= Solution
It is convenient to evaluate the <one-particle irreducible> vertices using the <Euclidean action> and then translate the <counterterms> back to the <Minkowski spacetime> <Lagrangian density>. For the <tadpole integral>, <analytic continuation> of <4/b/solution> gives
$$
I_{D,1}=\frac{\Gamma(-1+\epsilon/2)}{(4\pi)^{2-\epsilon/2}}(m^2)^{1-\epsilon/2}
=-\frac{m^2}{8\pi^2\epsilon}+O(1).
$$
Indeed $\Gamma(-1+\epsilon/2)=\Gamma(\epsilon/2)/(\epsilon/2-1)=-2/\epsilon+O(1)$. The negative pole is a property of <analytic continuation>, not the value of a convergent positive integral.
The <Euclidean space> inverse two-point <vertex> has the one-loop correction
$$
\Gamma_E^{(2)}(p_E)=p_E^2+m^2+\frac{\lambda_0}{2}I_{D,1}+O(\lambda_0^2),
\qquad
\left.\Gamma_E^{(2)}\right|_{\rm pole}=-\frac{\lambda_0m^2}{16\pi^2\epsilon}.
$$
The factor $1/2$ is the <symmetry factor> weight of the <tadpole diagram>. A mass <counterterm> $\delta m^2=\lambda_0m^2/(16\pi^2\epsilon)$ cancels this pole. There is no momentum-dependent pole at this order, and hence no one-loop <wave-function renormalization>.
For the four-point <vertex>, the three <bubble diagrams> give
$$
\Gamma_E^{(4)}=\lambda_0-\frac{\lambda_0^2}{2}(B_s+B_t+B_u)+O(\lambda_0^3),
\qquad
\left.\Gamma_E^{(4)}\right|_{\rm pole}=-\frac{3\lambda_0^2}{16\pi^2\epsilon}.
$$
One can check the relative minus sign independently from the one-loop <effective action>: its $\tfrac12\operatorname{Tr}\log(1+G\lambda_0\phi^2/2)$ term has a positive term linear in $\lambda_0\phi^2$ and a negative term quadratic in it. Thus $\delta\lambda=3\lambda_0^2/(16\pi^2\epsilon)$ cancels the quartic pole.
In the <Euclidean action>, these two <counterterms> are $+\delta m^2\phi^2/2+\delta\lambda\phi^4/4!$. Returning to the <Minkowski spacetime> <Lagrangian density> gives
$$
\boxed{\mathcal L_{\rm ct}=-\frac12\frac{\lambda_0m^2}{16\pi^2\epsilon}\phi^2
-\frac1{4!}\frac{3\lambda_0^2}{16\pi^2\epsilon}\phi^4.}
$$
For a direct sign check with <Feynman rules>, the two amputated loop pole insertions are $+i\lambda_0m^2/(16\pi^2\epsilon)$ and $+i3\lambda_0^2/(16\pi^2\epsilon)$, while the <counterterm> insertions are $-i\delta m^2$ and $-i\delta\lambda$.
As usual, the displayed pole residues suppress <renormalization scale> powers: $[\lambda_0]=\epsilon$ away from four dimensions. The next part restores these powers explicitly with a dimensionless <renormalized coupling>. Replacing the tree coupling by its renormalized counterpart inside a one-loop residue changes only higher-loop terms.
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