= Solution
At $s=0$ and $s=4T$, the mission is at the planet's position $x=c/\omega$. The final coordinate time from the last segment is
$$
t_{\mathrm{return}}=\frac4\omega\sinh(\omega T).
$$
The planet is at rest in these coordinates, so its <proper time> elapsed equals this coordinate-time difference. The astronauts' <proper time> is the parameter interval, $4T$. Therefore
$$
\boxed{\Delta\tau_{\mathrm{planet}}=\frac4\omega\sinh(\omega T),\qquad
\Delta\tau_{\mathrm{ship}}=4T.}
$$
For <proper acceleration> $c\omega=10\ \mathrm{m\,s^{-2}}$ and $c=3\times10^8\ \mathrm{m\,s^{-1}}$, $\omega^{-1}=3\times10^7$ seconds, approximately one year. Thus $\omega T=16$ and $e^{16}\approx3000^2=9\times10^6$. Neglecting $e^{-16}$ in the <hyperbolic sine>,
$$
\boxed{\Delta\tau_{\mathrm{planet}}\approx18\times10^6\text{ years},\qquad
\Delta\tau_{\mathrm{ship}}=64\text{ years}.}
$$
The clocks compare two different timelike paths between the same departure and reunion events.
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