13. In a football tournament with 10 teams, a double round-robin system is used, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point, and a loss earns 0 points.
(1) How many matches were played in total?
(2) If every match results in the maximum points, what is the total points accumulated by all teams? If every match results in the minimum points, what is the total points accumulated by all teams?
(3) If the teams are ranked by their points at the end of the tournament, what is the maximum point difference between two teams with adjacent rankings?
(1) Using a list or formula calculation, it can be known that a total of 90 matches were played in this competition.
(2) The highest score for a single match is 3 points. If the maximum score is achieved in every match, the total points of all teams in this competition would be $3 \times 90 = 270$ (points): The lowest score for a single match is 2 points. If the minimum score is achieved in every match, the total points of all teams in this competition would be $2 \times 90 = 180$ (points).
(3) Let the points difference between the $k$-th and $(k+1)$-th ranked teams be the maximum. Since the total score of each match is at most 3 points and at least 2 points, the maximum score for the top $k$ teams playing against each other is
$$
3 \times 2 \times \frac{k(k-1)}{2} = 3 k(k-1) \text{ (points); }
$$

The maximum score for the top $k$ teams playing against the remaining $10-k$ teams is
$$
3 \times 2 \times k(10-k) = 6 k(10-k) \text{ (points). }
$$

Therefore, the total score of the top $k$ teams is at most $3 k(19-k)$ points. Thus, the maximum score for the $k$-th ranked team is $3(19-k)$ points.

For the remaining $10-k$ teams, the minimum total score for their matches against each other is
$$
2 \times 2 \times \frac{(10-k)(9-k)}{2} = 2(10-k)(9-k) \text{ (points), }
$$

and their score from matches against the top $k$ teams is 0.
Therefore, the minimum score for the $(k+1)$-th ranked team is
$$
\frac{2(10-k)(9-k)}{10-k} = 2(9-k) \text{ (points). }
$$

Thus, when $k=1$, the point difference reaches its maximum value, with the two teams differing by $3(19-k) - 2(9-k) = 38$ points.