Answer
The relative speed of the two spaceships as measured by a passenger on either spaceship is $0.946~c$
Work Step by Step
Let $v_{MA}$ be the velocity of the moon relative to spaceship A. Then $v_{MA} = 0.80~c$
Then $v_{BM} = 0.60~c$
We can find $v_{BA}$:
$v_{BA} = \frac{V_{BM}~+~v_{MA}}{1+\frac{(v_{BM})~(v_{MA})}{c^2}}$
$v_{BA} = \frac{(0.60c)~+~(0.80c)}{1+\frac{(0.60c)~(0.80c)}{c^2}}$
$v_{BA} = \frac{1.40c}{1.48}$
$v_{BA} = 0.946~c$
The velocity of spaceship B relative to spaceship A is $0.946~c$
Therefore, the relative speed of the two spaceships as measured by a passenger on either spaceship is $0.946~c$
