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 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, and for 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 such that the requirement can be met in 5 weeks.
Since $66=18+12+12+12+12$, from the second week onwards, as long as each table generates 4 new edges.
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 can be as follows:
\begin{tabular}{|c|c|c|c|}
\hline Week Table & Table 1 & Table 2 & Table 3 \\
\hline One & $1,2,3,4$ & $5,6,7,8$ & $9,10,11,12$ \\
\hline Two & $1,2,11,12$ & $3,4,5,6$ & $7,8,9,10$ \\
\hline Three & $1,2,5,6$ & $7,8,11,12$ & $3,4,9,10$ \\
\hline Four & $1,2,7,8$ & $3,4,11,12$ & $5,6,9,10$ \\
\hline Five & $1,2,9,10$ & $3,4,7,8$ & $5,6,11,12$ \\
\hline
\end{tabular}

In conclusion, the minimum number of weeks required is 5.