Answer
(a) $0.50c$
(b) more than
(c) $0.73c$
Work Step by Step
(a) We know that
$v_{13}=\frac{v_{23}-v_{21}}{1-\frac{v_{23}v_{21}}{c^2}}$
We plug in the known values to obtain:
$v_{13}=\frac{0.80c-0.50c}{1-\frac{(0.80c)(0.50c)}{c^2}}$
$\implies v_{13}=0.50c$
(b) It is given that the speed of ship A relative to Earth $(v_{23})$ increases by $0.10c$ and therefore, the relative speed between ship A and ship B will be more than $0.10c$.
(c) We know that
$v_{21}=\frac{v_{23}-v_{13}}{1-\frac{v_{23}v_{13}}{c^2}}$
We plug in the known values to obtain:
$v_{21}=\frac{0.90c-0.50c}{1-\frac{(0.90c)(0.50c)}{c^2}}=0.73c$