(3) Let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it. Find the minimum value of $n$.
First, we prove that \( n \geqslant 11 \).

Consider a convex heptagon \( A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} \) with vertices in \( M \). Connect \( A_{1} A_{5} \). By condition (1), there is at least one point in \( M \) within the convex pentagon \( A_{1} A_{2} A_{3} A_{4} A_{5} \), denoted as \( P_{1} \). Connect \( P_{1} A_{1} \) and \( P_{1} A_{5} \). Then, in the convex pentagon \( A_{1} P_{1} A_{5} A_{6} A_{7} \), there is at least one point in \( M \), denoted as \( P_{2} \), and \( P_{2} \) is different from \( P_{1} \). Connect the line \( P_{1} P_{2} \). Among \( A_{1}, A_{2}, \cdots, A_{7} \), at least 5 vertices are not on the line \( P_{1} P_{2} \). By the pigeonhole principle, there must be 3 vertices on one side of the line \( P_{1} P_{2} \). These 3 vertices, together with points \( P_{1} \) and \( P_{2} \), form a convex pentagon that contains at least one point in \( M \), denoted as \( P_{3} \).

Next, draw the lines \( P_{1} P_{3} \) and \( P_{2} P_{3} \). Let the region \( \pi_{3} \) be the half-plane bounded by the line \( P_{1} P_{2} \) and on the opposite side of \( \triangle P_{1} P_{2} P_{3} \) (excluding the line \( P_{1} P_{2} \)). Similarly, define regions \( \pi_{1} \) and \( \pi_{2} \). These three regions \( \pi_{1}, \pi_{2}, \pi_{3} \) cover all points on the plane except for \( \triangle P_{1} P_{2} P_{3} \). By the pigeonhole principle, among the 7 vertices \( A_{1}, A_{2}, \cdots, A_{7} \), there must be \( \left[\frac{7}{3}\right] + 1 = 3 \) vertices in the same region (let's say \( \pi_{3} \)). These 3 points, together with \( P_{1} \) and \( P_{2} \), form a convex pentagon with vertices in \( M \), so its interior must contain at least one point in \( M \), denoted as \( P_{4} \). Therefore, \( n \geqslant 11 \).

Next, we construct an example to show that \( n = 11 \) is possible. As shown in Figure 1, the convex heptagon \( A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} \) is an integer-point heptagon with 4 points inside, so condition (1) is clearly satisfied. This point set \( M \) also satisfies condition (2). The proof is as follows: by contradiction, assume there exists an integer-point convex pentagon whose interior does not contain any integer points. Then, because the area of any integer-point polygon can be expressed as \( \frac{n}{2} \) (where \( n \in \mathbf{N}_{+} \)), by the principle of the smallest number, there must be a smallest area integer-point convex pentagon \( A B C D E \) whose interior does not contain any integer points. Considering the parity of the coordinates of the vertices, there are only 4 cases: (odd, even), (even, odd), (odd, odd), (even, even). Thus, among the vertices of the pentagon \( A B C D E \), there must be two vertices with the same parity for both coordinates, so their midpoint \( P \) is still an integer point. Since \( P \) is not inside the convex pentagon \( A B C D E \), \( P \) must be on one of the sides of the pentagon. Without loss of generality, assume \( P \) is on side \( A B \), then \( P \) is the midpoint of \( A B \). Connect \( P E \), then \( P B C D E \) is a smaller area integer-point convex pentagon whose interior does not contain any integer points, which is a contradiction.
In conclusion, the minimum value of \( n \) is 11.