How to find an angle in range(0, 360) between 2 vectors? I know that the common approach in order to find an angle is to calculate the dot product between 2 vectors and then calculate arcus cos of it. But in this solution I can get an angle only in the range(0, 180) degrees. What would be the proper way to get an angle in range of (0, 360)?

Before reading this answer - Imagine your angle in a 3D space - you can look at it from the "front" and from the "back" (front and back are defined by you). The angle from the front will be the opposite of the angle that you see from the back. So there is no real sense in a value in a range larger than $[0,180]$.
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In the 3D case, your two vectors would be on some plane (the plane that you can get its normal from the cross-product of the two vectors). Getting a $[0,180]$ degrees angle is of course possible by computing $arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$.
I think that what you can do, is to fix the Z axis, such that the two vectors will only differ in X and Y. Then you solve a 2D geometry problem. You should be able to fix the Z axis by dividing the two vectors by their Z component. Since the Z component is just a scalar, the vector directions will remain the same (the trend might change though, if this scalar is negative). You should remember whether these two scalars were both positive, both negative or one positive and one negative. In the last case, you're final results will be flipped!
If the Z component of the two vectors is $0$, then this step can (and actually must) be skipped. However, if only one of them has $0$ in its Z component, then it is probably impossible to compute the very precise angle, but you can still compute an approximation, by dividing the other vector by a very large number. 
Having that, you can reduce vector $\vec{a}$ from both vectors $\vec{a}$ and $\vec{b}$ and then add the vector $(1, 0, 0)$ to both. Thus, vector $\vec{a}$ will be $(1, 0, 0)$, while vector $\vec{b}$ will be $(b_x, b_y, 0)$.
Now, if $b_y$ is positive, than your angle is $arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$. Else, it is $360 - arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$. Remember to use the opposite result if only one of the original Z values was negative, as described above!
