Answer 1:
What you're asking is basically the essence of relativity: if you are sitting in a car that is going ALMOST the speed of light, then as far as you're concerned, you aren't moving, and if you throw a rock out of the car, then from your prospective, the rock will be moving exactly as fast as if you had been standing on the ground when you threw it instead. Somebody standing on the ground would see the rock moving at almost the speed of light as well, although just a little bit farther than your car. To calculate exactly how fast, do the following: Let v (lowercase) be the velocity of the car, u the velocity of the rock from the prospective of somebody in the car, and V (uppercase) the velocity of the rock as seen by somebody outside of the car. The rock has momentum p which is equal to the following formula: p = m * ((v / (1  (v^{2} / c^{2}))^{(1/2)}) + (u / (1  (u^{2} / c^{2}))^{(1/2)})) m = mass of the rock (which you don't need to know, because it drops out of the next step)c = speed of light.* is a multiplication sign (instead of x because x is often a variable). Also, p = m * (V / (1  (V^{2} / c^{2}))^{(1/2)})
Substitute the top expression for p, then do the algebra and solve for V. Then you can plug in your u and v and calculate V. Note: it's easier if you express u and v in proportion to the speed of light, for example u = 0.9c if u is 90% the speed of light (0.9c = 270,000 kilometers per second). This will make the numbers a LOT easier to keep track of. Also note: this illustrates WHY you can't go the speed of light  you would wind up dividing by zero.
