2. 12 friends agree to go out for a meal once a week, each time evenly divided into 3 tables. Each table seats 4 people, until any two people have eaten at the same table at least once. How many weeks are needed at a minimum?
2. Use 12 points $v_{1}, v_{2}, \cdots, v_{12}$ to represent 12 friends. When two friends sit at the same table for the first time in week $i$, a line labeled $i$ is drawn between the points representing these two friends, where $i=1,2, \ldots$. Thus, the problem is transformed into determining the minimum number of weeks required to form a complete graph $K_{12}$ in this manner.
It is known that the number of edges in $K_{12}$ is $C_{12}^{2}=66$.
In the first week, the number of edges labeled 1 is $3 \times C_{4}^{2}=18$.
In the second week, the 4 people who sat at the same table in the first week are seated at 3 different tables, so at least two of them will sit together, thus at least one edge labeled 2 is missing. With 3 tables, at least 3 edges are missing, so the number of edges labeled 2 is at most $18-3=15$.
Similarly, when $i \geqslant 3$, the number of edges labeled $i$ is at most 15.
Since $18+15+15+15=63<66$, the number of weeks required is $\geqslant 5$.
Next, we construct a seating arrangement that can achieve the required conditions in 5 weeks.
Since $66=18+12+12+12+12$, starting from the second week, it is sufficient to generate 4 new edges per table each week.
Divide the 12 people into 6 pairs, with the same pair always sitting at the same table, and different pairs sitting together exactly once in 5 weeks. The seating arrangement is as follows:
In summary, the minimum number of weeks required is 5.
3. Let $1,2, \cdots, n$ be $n$ vertices, and connect an edge between $a$ and $b$ if and only if $(a, b) \in S$, resulting in a simple graph $G=(V, E)$.
Thus, the original problem is transformed into: Given a graph $G$ with $n$ vertices and $m$ edges, the graph contains at least $\frac{4 m}{3 n}\left(m-\frac{n^{2}}{4}\right)$ triangles.
Let the degree of $V_{i}$ be $d\left(V_{i}\right)$, and we estimate the number of triangles in $G$:
The number of triangles containing an edge $\left(V_{i}, V_{j}\right)$ in $G$ is at least $d\left(V_{i}\right)+d\left(V_{j}\right)-n$,
Since each triangle in $G$ is counted three times, the number of triangles in $G$ is at least
$$
\begin{array}{l}
\frac{1}{3} \sum_{\left(V_{i} \cdot V_{j}\right) \in E}\left[d\left(V_{i}\right)+d\left(V_{j}\right)-n\right]=\frac{1}{3}\left\{\sum_{\left(V_{1} \cdot V_{j}\right) \in E}\left[d\left(V_{i}\right)+d\left(V_{j}\right)\right]-m n\right\} \\
=\frac{1}{3}\left\{\sum_{i=1}^{n}\left[d\left(V_{i}\right)\right]^{2}-m n\right\} \geqslant \frac{1}{3}\left\{\frac{\left[\sum_{i=1}^{n} d\left(V_{i}\right)\right]^{2}}{n}-m n\right\}=\frac{1}{3}\left(\frac{4 m^{2}}{n}-m n\right)=\frac{4 m}{3 n}\left(m-\frac{n^{2}}{4}\right) .
\end{array}
$$

Thus, the conclusion holds.