Your approach to the first question is entirely reasonable. Since $5\cdot6=30>28$, it’s clear that $6$ wins guarantee advancement. To show that $5$ do not, we need only show that it’s possible for $5$ teams to win $5$ matches each; the arrangement shown by user73985 works fine and is natural enough that it’s the first one that I found as well. (You can complete it by distributing the remaining $3$ wins within the group consisting of teams $6,7$, and $8$; for example, you could have team $6$ beat teams $7$ and $8$, and team $7$ beat team $8$.)
For the second question, suppose that a team has $3$ wins; is it possible that it could advance to the second round? Suppose that teams $1,2$, and $3$ beat each of the higher-numbered teams: team $1$ beats everybody else, team $2$ beats everyone else except team $1$, and team $3$ beats everyone else except teams $1$ and $2$. That accounts for $18$ wins, leaving $10$ to be determined. Team $4$ beats teams $5,6$, and $7$, for $3$ wins, leaving $7$ wins to still to be distributed amongst teams $5,6,7$, and $8$. We want team $4$, with its measly $3$ wins, to make it to the second round, so we’ll try to split the remaining $7$ wins $2,2,2$, and $1$ amongst teams $5,6,7$, and $8$. Team $4$ did not beat team $8$, so team $8$ beat team $4$ and already has one win; we can give it another against team $5$. Now it must lose to teams $6$ and $7$, so they now have one win apiece. We can finish up by letting team $5$ beat team $6$, team $6$ beat team $7$, and team $7$ beat team $5$. To sum up:
- Team $1$ beats teams $2,3,4,5,6,7,8$.
- Team $2$ beats teams $3,4,5,6,7,8$.
- Team $3$ beats teams $4,5,6,7,8$.
- Team $4$ beats teams $5,6,7$.
- Team $5$ beats team $6$.
- Team $6$ beats teams $7,8$.
- Team $7$ beats teams $5,8$.
- Team $8$ beats teams $4,5$.
Thus, it’s possible to make it into the second round with just $3$ wins.
With only $2$ wins, however, it’s impossible to guarantee getting to the second round. The top three finishers cannot have more than $18$ wins altogether (either $7+6+5$ or $6+6+6$). That leaves $10$ wins amongst the remaining $5$ teams, so either each of the bottom $5$ teams has $2$ wins, or one of them has at least $3$ wins. In neither case is a team with just $2$ wins assured of getting into the second round.
There is some ambiguity in the situation in which each of the bottom $5$ teams wins $2$ matches. (This can happen, e.g., if teams $1,2$ and $3$ all beat each of teams $4,5,6,7$, and $8$, team $4$ beats teams $5$ and $6$, team $5$ beats teams $6$ and $7$, team $6$ beats teams $7$ and $8$, team $7$ beats teams $8$ and $4$, and team $8$ beats teams $4$ and $5$.) In this case the rules as given in the question don’t specify what happens. Some of the possibilities are: that only the top three teams go to the second round; that there is a playoff for the fourth position; that the fourth position is chosen randomly; or that the fourth position is decided by some tie-breaker like goal differential. As long as four teams always move on to the second round, it’s still possible for a team with just $2$ wins in the first round to move on, but if that does happen, other teams with $2$ wins fail to move on.
Best Answer
There's a counterexample with 8 teams: A, B, C, X, Y, Z, M, S
Teams ABCXYZ each win twice and lose twice for a score of $2\times0+3\times1+2\times3=9$.
Team M wins against ABC and loses to XYZS, for a score of $4\times0+0\times1+3\times3=9$.
Team S wins once and ties 6 times, for a score of $0\times 0+6\times1+1\times3=9$.
How I found this: First, for easier counting I changed the rules by subtracting $n-1$ points from each team such that ties give $0$ points, and wins and losses each $2$ and $-1$. That makes it easier to see which possible combinations of wins and losses add up to the same. Then I'm looking for a directed graph with no 2-loops such that the value of each node is the same.
Since there are as many wins as losses, in a counterexample there must be at least one team that win more times than they lose. But that team cannot possibly have less than $2$ points, so let's see which way we can make a node with value 2: They are: $1W+0L$, $2W+2L$, $3W+4L$ and so forth -- so if we have one of $1W+0L$ and $3W+4L$ the number of wins/losses add up right.
From there it was just a matter of puzzling out where to add $2W+2L$ nodes to the graph such that we don't need more than one match to take place between the same two teams.