All jokes aside (see comment on OP)...
Let the people be numbered $P_1, P_2, P_3,\ldots$ and the books $B_1, B_2, B_3, \ldots$
If we have $P_n$ start on book $B_n$, the next book they have not yet read is book $B_{n+1}$. If this book does not exist, wrap around to the start of the shelf.
Essentially, have each person start with a different book, and move down the shelf one-by-one.
No unnecessary books are opened, so this is optimal.
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
Addendum added to respond to the comment/question of Arty.
Thanks to Ross Millikan for pointing out:
The obvious mistake of my confusing the total number of points scored with the total number of points scored by all of the girls.
The more subtle mistake that the average score of all of the girls must not be greater than what their score would be, if every girl beat every boy.
If $x$ girls scored $P$ points, then $3x$ boys scored $1.2P$ points, so $2.2P$ points were scored in total.
Let $~Q = 2.2P = ~$ be the total number of points scored.
Then, the total score of all the boys is $~\dfrac{6Q}{11}~$ and the total score of all the girls is $~\dfrac{5Q}{11},~$ so $~Q~$ must be a multiple of $~11.~$
Also, each game generated $1$ point, distributed as either [1,0] or [1/2,1/2]. Therefore $~Q~ = ~$ the number of games played.
Since there were $~x~$ girls and $~3x~$ boys, you have that
$$Q = \binom{4x}{2} = (2x) \times (4x - 1).$$
So, $~Q~$ must be chosen so that there exists a positive integer $x$ such that
$(2x) \times (4x-1) = Q.$
So, either $2x$ is a multiple of $11$ or $(4x - 1)$ is a multiple of $11$.
The smallest positive integer $x$ that can serve as a $\color{red}{\text{candidate value}}$ is $x = 3.$
Exploring this $\color{red}{\text{candidate value}}$:
The total number of games played is
$\displaystyle \binom{12}{2}~ = 66 = Q.$
The total number of girls is $~3~$ and the total number of boys is $~9.$
The total number of points collectively scored by all of the girls is $~\dfrac{5Q}{11} = 30.$
So, the average score of each girl was $~10.$
The total number of points collectively scored by all of the boys is $~\dfrac{6Q}{11} = 36.$
So, the average score of each boy was $~4.$
$\color{red}{\text{Does this work?}}$
Suppose that every girl beat all 9 boys.
Further suppose that every girl:girl game and every boy:boy game ended in a draw.
Then each girl would score
$\displaystyle 9 + \left[2 \times \frac{1}{2}\right] = 10,$
and each boy would score
$\displaystyle 0 + \left[8 \times \frac{1}{2}\right] = 4.$
So, the $\color{red}{\text{candidate value}}$ of $x = 3$ works okay.
Addendum
Responding to the comment/question of Arty.
First of all, your overall approach was $\color{red}{\text{better}}$ than my $\color{red}{\text{eventual}}$ approach, in one way, and flawed in another way.
That is:
You elegantly confronted the issue that the total score of the girls, assuming that each girl beat each boy, had to be at least $~\dfrac{5}{11}~$ of the total games played.
Initially, I totally overlooked that issue. Then, after Ross Millikan commented, I edited my answer after the fact. However, I did not try for the elegant (and better) approach of establishing an inequality involving $~x.~$
Instead, I (inelegantly) used the notion of a candidate value to verify that my solution worked.
You totally overlooked the constraint that the number of games played, $~\dfrac{4x(4x-1)}{2}~$ has to be a multiple of $~11.~$ This is because the total number of points scored by the girls is
$\displaystyle y = \frac{5}{11} \times \frac{4x(4x-1)}{2}.$
I agree with your updated interpretation of the above inequality. That is you have that:
$\displaystyle \frac{40x^2 - 10x}{11} \leq 3x^2 + \frac{x^2 - x}{2} = \frac{7x^2 - x}{2}.$
Cross multiplying, this implies that
$\displaystyle 80x^2 - 20x \leq 77x^2 - 11x.$
Since $x$ must be positive, (so $x$ is non-zero), you can divide the above inequality through by $x$ to give
$80x - 20 \leq 77x - 11 \implies 3x \leq 9 \implies x \leq 3.$
This begs the question, which value of $x$ is satisfactory.
You need $~\displaystyle \frac{4x(4x-1)}{2} = (2x) \times (4x-1)~$ to be a multiple of $11$.
Of the candidate values given by $~x \in \{1,2,3\}~$, only $~x=3~$ satisfies this constraint.