An observer on Earth watches two spaceships (also with observers) fly by, one coming from the left, one coming from the right. (The space ships are on parallel paths and will pass each other closely but won't collide.) Each spaceship is going 75% the speed of light with respect to the Earth observer.
What does each observer see, regarding the other two?
Your question is not clear:
1- If you want to calculate the speed of one of the spaceships from the viewpoint of the other one, you have to use relativistic velocity-addition formula:
$$w=\frac{u+v}{1+uv/c^2}$$
If the velocity of the spaceships are similar and $u=v=0.75c$, thus you have:
$$w=\frac{0.75+0.75}{1+0.75^2}c=0.96c$$
2- If you want to calculate the virtual velocity of the contracting distance between the spaceships due to their motions towards each other from the viewpoint of the Earth observer, you have to calculate $0.75c+0.75c=1.5c$ (Remember that the velocity of a contracting/expanding distance, which is not a real object, is allowed to be greater than the speed of light in the framework of special relativity. This velocity can reach a maximum value $2c$ from the viewpoint of any inertial observer, say the velocity of the expanding distance between two receding photons from the viewpoint of any observer)
