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.
Maximum number of wins with which a team may not qualify: $10$
The idea here is to list all the teams in order of their number of wins, and maximize the $5$th best team's wins.
Hence we want as many total wins as possible for the top $5$ teams - meaning we want the least total possible wins for the bottom three teams. What is this least number? Well each of these three teams must play each other in a total of $6$ games, so together they account for a minimum of $6$ wins. The total number of games is $56$, so the top $5$ teams get to distribute $56-6=50$ wins. Now the best that the $5$th best team can therefore achieve is $\frac{50}{5}=10$ wins. Note that this is possible if each of the top $5$ teams always beats the bottom $3$ teams ($6$ wins) and then wins exactly one match against the other top teams ($4$ wins) - giving a total of $10$ wins to the $5$-th best team. Hence $10$ is the most number of wins which does not guarantee qualification (though this is an extremely rare case).
Minimum number of wins to qualify: $4$
The idea here is to apply the same procedure and look at the performance of the $4$-th best team. Each of the bottom $5$ teams must play each other - giving them $20$ total victories to distribute. Hence the $4$-th best person must have at minimum $\frac{20}{5}=4$ victories. This is again achievable - each of the bottom $5$ teams wins exactly once against each other and looses all other matches.
Best Answer
There are $42$ points available.
1) The top team can win all games $(12 pts)$ and the second team losing one game $(10 pts)$. The other $5$ teams could all draw with each other so there would be $5$ teams in $3rd$ place with $4$ points each.
2) The $6th$ and $7th$ placed teams could lose all their matches except when they play each other so they would share $2$ points $(W1, D0, L5)$ and $(W0, D0, L6)$. This leaves $40$ available points between $5$ teams. The top team could get $9$ points $(W4, D1, L1)$, the next $3$ teams $8$ points each $(W3, D2, L1)$ and the $5th$ placed team $7$ points $(W3, D1, L2)$. These win, draw and lose metrics are compatable.