Assuming there is no order requirement:
Since the numbers 1-8 already appear twice inside the sixteen given numbers,
we just need to choose an extra string 1-8, and nothing more. Then, the third
property is a consequence of the last two (and hence, dependent on them), as it
always creates four numbers which are common and four numbers which are not
common between any two strings of (the eventual) six numbers.
After a quick look, it seems there are more than one solutions (I think fifteen),
as depending on the choice of first two strings, we end up with two "searching
possibilities" with either three or two solutions). For example, here's two of them:
I would suggest imposing a third condition for uniqueness, such as relating
their ascending order.