How many subsets are there such that their pairwise intersection is empty? Let $A$ be a set of $n$ numbers, I need to find the max collection of non empty subset of $A$ such that intersection of any two pair is empty.

To me answer seem $n$, take the example $A = \{1,2,3\}$, there are only three non-empty subsets ({1},{2},{3}) which satisfy the required condition.

In general what is the maximum number of subsets in a collection of subsets such that intersection of any two pair is at most $k$?
No element of $A$ can be in two of the subsets, or it would be in the intersection of that pair.  The approach you hint at, taking all the singletons, is optimal.
I think the answer is $\sum_{i=1}^{k+1} \binom {n}{i}$ but a reasonable proof eludes me.
The sum counts all the nonempty subsets of size $ \le k+1$. The intersection of any two of them that are distinct can have no more than $k$ elements in common.  So the desired number is $\ge$ $\sum_{i=1}^{k+1} \binom {n}{i}$ 

Going the other way is harder. A collection larger than $\sum_{i=1}^{k+1} \binom {n}{i}$ must contain a subset of size $\ge k+2$. If a set of size $h: n\ge h \ge k+2$ is included then $ \binom{h}{k+1} $ sets of size $k+1$ must be excluded. Applying this process one should be able to show that any other collection of non-empty subsets with pairwise intersections of size $\le k$ must contain fewer subsets then $\sum_{i=1}^{k+1} \binom {n}{i}$.