= Solution
The separation simultaneous in the mothership frame at launch is the <length contraction> of the Earth-frame simultaneous separation:
$$
D'=\frac{2\,\mathrm{ly}}{\gamma_m}=1.60\,\mathrm{ly},\qquad\gamma_m=\frac1{\sqrt{1-0.6^2}}=1.25.
$$
In that frame Earth approaches the launch position at $0.6c$ while the lander travels toward Earth at $0.8c$. Their coordinate separation closes at $1.4c$, so
$$
\boxed{t'_{\rm arrival}=\frac{1.60\,\mathrm{ly}}{1.4c}=\frac87\,\mathrm{yr}=1.14\,\mathrm{yr}.}
$$
A closing rate between two objects can exceed $c$; it is not either object's speed in the other's rest frame. Simultaneity matters: simply transforming the Earth event at Earth-frame launch time would not give the simultaneous mothership-frame separation.
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