= Solution
Put $a=\omega T$. At emission, $s=2T$, the middle segment gives
$$
t_e=\frac2\omega\sinh a,\qquad x_e=\frac c\omega(2\cosh a-1).
$$
The planet is at $x_p=c/\omega$, so the inward <light> signal takes coordinate time
$$
\frac{x_e-x_p}{c}=\frac2\omega(\cosh a-1).
$$
Its arrival time on the planet is therefore $t_r=2(e^a-1)/\omega$. Subtract this from the mission's return time $4\sinh a/\omega$:
$$
\boxed{t_{\mathrm{return}}-t_r=\frac2\omega(1-e^{-\omega T}).}
$$
For large $T$ at fixed acceleration this tends to $2/\omega=2c/(c\omega)$. With the Earth-gravity value above, the wait is \b[approximately two years], despite the millions of planet-years elapsed since departure in part (i).
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2014-accelerated-journey.png]
{title=An accelerated round trip and the midpoint signal received before reunion}
{description=Original spacetime diagram at omega T=1.2 for visibility. The planet follows the vertical line. The mission sends an inward light signal from its most distant point; the signal reaches the planet before the final reunion. Both axes use the same dimensionless spacetime scale.}
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