= Solution
{c}
The <Earth>-<Moon> pair is a <gravitationally bound system>, so its separation does not obey $\dot r=H_0r$. The <Hubble flow> describes the motion of the approximately homogeneous cosmological matter distribution; it is not a compulsory extra velocity assigned to the members of every local orbit. Thus the numerical proximity of the counterfactual rate to $3.8\,\mathrm{cm\,yr^{-1}}$ is not evidence for that explanation.
The observed recession arises principally from <tidal dissipation>. The <Earth> rotates faster than the <Moon> orbits, so the displaced tidal response transfers <angular momentum> from the <Earth>'s spin into the lunar orbit. For a nearly circular orbit, increasing orbital <angular momentum> increases its size. A model of this effect must include the appropriate tidal response, rather than add a universal <Hubble flow> recession.
A residual <local cosmological tidal acceleration> can exist, but its background scale is $|(\ddot a/a)r|\sim H_0^2r$, not an imposed first-order secular drift $H_0r$. At $r=4\times10^8\,\mathrm m$,
$$
H_0^2r\sim2\times10^{-27}\,\mathrm{m\,s^{-2}},
\qquad
\frac{G(M_E+M_M)}{r^2}\sim2.5\times10^{-3}\,\mathrm{m\,s^{-2}},
$$
using $G(M_E+M_M)\simeq4\times10^{14}\,\mathrm{m^3\,s^{-2}}$. The cosmological scale is about $10^{-24}$ of the mutual <gravitational acceleration>. Its detailed coefficient depends on how the local system is embedded in the cosmological background. \b[Cosmological expansion can be neglected for the lunar recession calculation at the stated precision]; this conclusion follows from binding and the tiny residual <local cosmological tidal acceleration>, rather than from comparing two recession speeds.
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