Question: Is there $A \subset \mathbb{R}^n$, such that $A$ is compact, contains the origin and has measure zero?
I know that the finite set A, of the form $A=\{ x\_{1},...,x\_{n} \}$, is a set of measure zero, but I don't know about its compactness nor if one can include the element zero amoung its elements.

Answer: 
A non-trivial one would be the standard Cantor set in $\mathbb{R}^n$. It's closed, bounded (contained in the unit rectangle $[0,1]^n$, thus compact. It also contains the origin and is of measure zero.
Whatever definition you use of compactness and of *measure zero*, the easiest example of $A \subset \mathbb{R}^n$ with all the properties you like is $A = \{0\}$.

No matter if you use sequential compactness or compactness via open coverings, a singleton is trivially compact and of null Lebesgue measure.