= Solution
Use renormalized <propagators> of <mass> $m$ and tree vertex $-ig\mu^\epsilon$. The interaction <counterterms> have Feynman factors $-i\delta_{m^2}$ for a two-leg <mass> insertion and $-i\mu^\epsilon g\delta_g$ for a four-leg insertion. At one loop the two-point <one-particle-irreducible vertex> is
$$
i\Gamma^{(2)}(p)=i(p^2-m^2)-i\delta_{m^2}
+(-ig\mu^\epsilon)\int_k\frac{i}{k^2-m^2+i0}+O(g^2).
$$
The directed tadpole has $S=1$: interchanging its two ends would reverse the <field> labels, so the real-scalar tadpole's factor $1/2$ does not apply. It is independent of external <momentum>, so no wave-function <counterterm> is required at this order.
For the four-point vertex, label the two unstarred incoming <fields> by $1,2$ and the starred ones by $3,4$, with all <momenta> incoming. The tree contact, the three oriented bubbles and the contact <counterterm> give
$$
\begin{aligned}
i\Gamma^{(4)}={}&-ig\mu^\epsilon-i\mu^\epsilon g\delta_g\\
&+(-ig\mu^\epsilon)^2\left[\frac12 I_2(p_1+p_2)+I_2(p_1+p_3)+I_2(p_1+p_4)\right]+O(g^3).
\end{aligned}
$$
The first bubble has two equivalent arrows along the same direction between its vertices, giving $S=2$. In the other two channels the arrows oppose one another, and interchanging the lines does not preserve their oriented attachments, giving $S=1$. This is the <oriented complex-scalar quartic graph factors> rule.
The <mass> tadpole is order $gm^2$, while the four-point loop is order $g^2$. In the multiplicative <coupling> convention of the source,
$$
\boxed{\delta_{m^2}=O(gm^2/\epsilon),\qquad \delta_g=O(g/\epsilon),\qquad g\delta_g=O(g^2/\epsilon).}
$$
The distinction between $\delta_g$ and the whole additive four-leg coefficient $g\delta_g$ matters for the requested order counting.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2017-complex-scalar-renormalization-graphs.png]
{title=Tree, loop and <counterterm> contributions to the two-point and four-point vertices of a <complex scalar field>}
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