What is escape velocity?
According to Wikipedia http://en.wikipedia.org/wiki/Earth
Earth:
Escape velocity: 11.186km/s?
What does that mean?
Does the projectile should be fired perpendicular with respect to the ground angle?

According to this:

According to this, can the projectile be fired in any direction (because it is speed not velocity) as long as its speed is 11.186 Km/s to escape earth?

If (supposed Earth has no atmosphere) the projectile is fired parallely to the ground (tangent), what is its escape velocity?

The wikipedia article has it right. As long as the trajectory doesn't intersect the surface of the earth and go "boom!", if it is launched at or above escape velocity the projectile will escape no matter what direction it is aimed. Escape velocity tangent to the surface of an airless non-rotating body is the same as escape velocity vertical to the surface.

There are many ways of showing this, but the easiest is to look at the sum of the projectile's potential and kinetic energy, which must remain constant throughout its trajectory.

Yes, yes, I understand now. So if the projectile is placed, say..., 1000m above ground (earth).
And it's fired horizontally and 1^{0} to the ground, supposed the earth has no atmosphere and no mountain (and skyscraper for that matter), I don't know if the math is correct for 1^{0}; 1000 m above the ground; and earth radius. After all earth is bolate not sphere.
Just supposed in that height and direction the projectile is fired 11.186 km/s, and doesn't go boom, it will travel to space, Right?