a. If all vertices of resulting triangles coincide with vertices of the heptagon, the sum of the angles of the triangles equals the sum of the angles of the heptagon and equals $5 \cdot 180^\circ$. Hence, there are only 5 triangles. In case if two diagonals intersect not at the vertex of the heptagon, then this intersection point $A$ lies inside the heptagon because of its convexity. Then the angles of triangles adjacent to point $A$ sum up to $360^\circ$, and the sum of the angles of resulting triangles is not less than $5 \cdot 180^\circ + 360^\circ = 7 \cdot 180^\circ$. Thus, in this case will have at least 7 triangles.

b. The heptagon and corresponding division are shown in fig. 38. Note that the horizontal diagonal lies through the vertex of the heptagon.