a) Clearly, there are 24 blue vertices. If all red vertices form a single block then $A = 78$, if they form two blocks then $A = 77$, and similarly, if they form $k$ blocks then $A = 79 - k$.

Besides, the number of red blocks is equal to the number of blue blocks, hence $B = 24 - k$. This implies that all possible values of $(A, B)$ are $(79 - k, 24 - k)$ for $k = 1, 2, \dots, 24$.

b) In order to have $B = 14$ we need $k = 10$, i.e. blue vertices form 10 blocks (and red vertices also form 10 blocks). Starting from a certain blue block, we label all blue blocks clockwise from 1 to 10. Let the numbers of vertices in blue blocks be $x_1, x_2, \dots, x_{10}$, respectively.

Let $y_i = x_1 + \dots + x_i$ then $1 \le y_1 < y_2 < \dots < y_9 < y_{10} = 24$. Hence, there are $\binom{23}{9}$ possible ways to choose such $y_1, \dots, y_{10}$. In other words, 24 blue vertices can form 10 blocks in $\binom{23}{9}$ possible ways. Similarly, red vertices can form 10 blocks in $\binom{78}{9}$ ways.

Hence, there are $\binom{23}{9}\binom{78}{9}$ possible ways to arrange 10 blue blocks and 10 red blocks alternating. Note that such arrangements are distinct by the rotation because 79 is a prime number. However, there are 10 ways to choose the starting block, hence the number of pairwise non-similar colorings is $\frac{\binom{23}{9}\binom{78}{9}}{10}$.

$\square$