= Solution
{c}
Use the <Earth>'s <inertial frame>, with the <Earth> at $x=0$ and the star at $x=D=4.2$ light years, neglecting their relative motion. Take the stated relative <velocity> of $B$ to be forward along the same direction as $A$. The <relativistic velocity addition> law gives
$$
\boxed{v_B=\frac{0.25c+0.50c}{1+(0.25)(0.50)}=\frac23c.}
$$
In units of light years and years, $c=1$ and the spacecraft <worldlines> are $x_A=t/4$ and $x_B=2t/3$. Their passages at the star occur at
$$
t_{B*}=\frac{4.2}{2/3}=6.3\,\mathrm{yr},\qquad
t_{A*}=\frac{4.2}{1/4}=16.8\,\mathrm{yr}.
$$
The return <radio wave> from $B$ follows the leftward light <worldline>
$$
x=4.2-(t-6.3)=10.5-t.
$$
Intersecting this with $A$'s <worldline> gives $t/4=10.5-t$, hence
$$
\boxed{t_{A\leftarrow B}=8.4\,\mathrm{yr},\qquad x_{A\leftarrow B}=2.1\,\text{light years}.}
$$
This is the <Earth>-frame coordinate time of reception on $A$, as requested; it is not the reading of $A$'s own clock. If an observer on the <Earth> watched that reception through a returning light signal, they would see it later.
Each transmission from the star takes another $D/c=4.2$ years to reach the <Earth>. Therefore
$$
\boxed{t_{\oplus\leftarrow B}=10.5\,\mathrm{yr},\qquad
t_{\oplus\leftarrow A}=21.0\,\mathrm{yr}.}
$$
All times are measured from the common passage by the <Earth>; the spacecraft continue at their stated <velocities> during the calculation.
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