a) Note that every polygon produced during this process is convex. That said, the idea here is to find an upper bound for the sum of angles of them. After cutting a polygon $P$ into two polygons $P_1$ and $P_2$, three possibilities arise: $(\sigma(Q)$ denotes the sum of angles of polygon $Q)$
* The cutting line connects two vertices of $P$. In this case $\sigma(P) = \sigma(P_1) + \sigma(P_2)$.
* The cutting line connects a vertex of $P$ to an inner point of a side. In this case $\sigma(P_1) + \sigma(P_2) = \sigma(P) + \pi$.
* The cutting line connects two inner points from two sides of $P$. In this case $\sigma(P_1) + \sigma(P_2) = \sigma(P) + 2\pi$.
The first polygon is a triangle and the sum of its angles is $\pi$. Therefore, after the first cut the sum of angles of the resulting polygons is at most $\pi + 2\pi = 3\pi$. After the second cut each of these two polygons are divided into two new polygons. Therefore, after the second cut the sum of angles of these polygons is at most $3\pi + 2 \times 2\pi = 7\pi$. Similarly, after the third cut the sum of angles of the resulting eight polygons is at most $7\pi + 4 \times 2\pi = 15\pi$. Now, if the 7 equal pieces have at least 5 sides, then the sum of angles of each polygon is at least $3\pi$ and the sum of angles of the last piece is at least $\pi$. Therefore,
$$22\pi = \pi + 7 \times 3\pi \le \text{sum of angles of all pieces} \le 15\pi.$$
This contradiction shows that these seven equal pieces must be either triangles or quadrilaterals.

b) According to the previous part, if these equal pieces are quadrilaterals, then the last piece should be a triangle. Furthermore, the cutting line of each polygon connects two inner points from two sides of the polygon. Assuming that such a cutting process is possible, a result is that the seven congruent quadrilaterals cannot be cyclic; because if they are cyclic, one of the following cases happens, and it is easy to show that each case results in a contradiction.

Assuming that the quadrilaterals are not cyclic, it is easy to show that they must be parallelograms. Like before, it is easy to check the resulting case and verify that it is impossible.

c) Yes! Let $x$ be a very small positive number. Then the cutting process can be done as instructed below: