7.1. Can several diagonals be drawn in a convex hexagon so that each diagonal intersects exactly three other diagonals inside the hexagon?
7.1. Not possible.

In a convex hexagon, there are two types of diagonals: one type is the "long diagonals," which connect two opposite vertices; the other type is the "short diagonals," which connect vertices that are one apart. If only long diagonals are drawn, the required property in the problem cannot be satisfied. Therefore, among the diagonals drawn, there must be short diagonals. However, once a short diagonal is drawn, for example, $A_{1} A_{3}$ (Figure 3), it can only intersect with the three diagonals drawn from $A_{2}$ inside the shape. 

Therefore, at this point, it is necessary to draw the short diagonal $A_{2} A_{4}$ and the long diagonal $A_{2} A_{5}$. Further examination of the short diagonal $A_{2} A_{4}$ reveals that it is also necessary to draw the short diagonal $A_{3} A_{5}$ and the long diagonal $A_{3} A_{6}$. Continuing this process, it becomes clear that all diagonals need to be drawn. At this point, it is obviously impossible to meet the requirements of the problem.