= Solution
Write
$$
I_j=\int_{\Lambda/\zeta<|k|<\Lambda}
\frac{d^dk}{(2\pi)^d}\,G_0(k)^j,
\qquad
G_0(k)=\frac1{k^2+\mu_0^2},
$$
and, for the three-line topology,
$$
J_3=\int_+\frac{d^dk\,d^dq}{(2\pi)^{2d}}
G_0(k)G_0(q)G_0(k+q),
$$
and define the four-line integral
$$
J_4=\int_+\frac{d^dk\,d^dq\,d^d\ell}{(2\pi)^{3d}}
G_0(k)G_0(q)G_0(\ell)G_0(k+q+\ell),
$$
with every propagator momentum restricted to the fast shell.
The first cumulant contains
$$
\lambda_0\binom62\langle(\phi^+)^4\rangle_+(\phi^-)^2
=45\lambda_0 I_1^2(\phi^-)^2.
$$
Since the quadratic free energy is $\frac12\mu^2(\phi^-)^2$, this gives
$$
\delta\mu^2\big|_{\lambda_0}=90\lambda_0I_1^2.
$$
The mixed term in the second cumulant is
$$
-\left(\langle V_6V_4\rangle_+
-\langle V_6\rangle_+\langle V_4\rangle_+\right).
$$
Wick contraction with the printed normalization $V_6=\lambda_0\int\phi^6$ and $V_4=g_0\int\phi^4$ gives, at zero external momentum,
$$
\delta\mu^2\big|_{\lambda_0g_0}
=-2\lambda_0g_0
\left(1620I_1^2I_2+1440I_1J_3+360J_4\right).
$$
The coefficients respectively combine the two placements of both external legs in the two-line topology, the three-line topology with one external leg on each vertex, and the four-line topology. Couplings normalized as $g_0\phi^4/4!$ and $\lambda_0\phi^6/6!$ absorb the corresponding factorials, which is why formulas in that convention have much smaller numerical coefficients.
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