tournament of 8 people in teams of two, everyone teaming up once and competing twice? We have a tournament of 8 people in changing teams of two. 14 games, held at two fields.
f.e.: Field One, Game One: AB vs CD Field Two, Game Two: EF vs GH
Is it possible to mix and team the 8 players so that they each play with everyone else once and against everyone else twice?
 A: Yes.  In effect, you are trying to create the complete graph on 8 vertices 3 different times.  First, we will address the matter of getting everyone to play with everyone else:
Let each person be a vertex on $K_8$.  Let two vertices be adjacent if the two corresponding players played together.  Note that in each of the 14 games, two new edges are made, one for each team.  This means that we have 28 edges in our graph on 8 vertices by the end of the tournament.  The size of $K_8$ is $\frac{7\cdot8}{2}=28$, which means that the tournament can fill out the entire graph.  This happens if each player plays in half the games, so their vertex has degree 7.
Note that combinatorics does not discriminate based on whether or not you are on a team with a person, so by a symmetry argument we observe that it is, indeed, possible for each person to play against each other person two times (if we arrange the players in a $2\times 2$ grid, each player can play each other player once while directly opposing each other and once while diagonal to each other).
