(a) The minimum is 0, which is achieved by a tournament in which team $T_i$ beats $T_j$ if and only if $i > j$.

(b) Any set of three teams constitutes either a cycle triplet or a "dominated triplet" in which one team beats the other two; let there be $c$ of the former and $d$ of the latter. Then $c + d = \binom{2n+1}{3}$. Suppose that team $T_i$ beats $x_i$ other teams; then it is the winning team in exactly $\binom{x_i}{2}$ dominated triples. Observe that $\sum_{i=1}^{2n+1} x_i = \binom{2n+1}{2}$, the total number of games. Hence

$$d = \sum_{i=1}^{2n+1} \binom{x_i}{2} = \frac{1}{2} \sum_{i=1}^{2n+1} x_i^2 - \frac{1}{2} \binom{2n+1}{2}$$

By the Cauchy-Schwarz Inequality, $(2n+1) \sum_{i=1}^{2n+1} x_i^2 \geq \left( \sum_{i=1}^{2n+1} x_i \right)^2 = n^2 (2n+1)^2$, whence

$$c = \binom{2n+1}{3} - \sum_{i=1}^{2n+1} \binom{x_i}{2} \leq \binom{2n+1}{3} - \frac{n^2 (2n+1)}{2} + \frac{1}{2} \binom{2n+1}{2} = \frac{n(n+1)(2n+1)}{6}.$$

To realize the upper bound, let the teams be $T_1 = T_{2n+2}, T_2 = T_{2n+3}, \cdots, T_i = T_{2n+1+i}, \cdots, T_{2n+1} = T_{4n+2}$. For each $i$, let team $T_i$ beat $T_{i+1}, T_{i+2}, \cdots, T_{i+n}$ and lose to $T_{i+n+1}, \cdots, T_{i+2n}$. We need to check that this is a consistent assignment of wins and losses, since the result for each pair of teams is defined twice. This can be seen by noting that $(2n+1+i) - (i+j) = 2n+1-j \geq n+1$ for $1 \leq j \leq n$. The cycle triplets are $(T_i, T_{i+j}, T_{i+j+k})$ where $1 \leq j \leq n$ and $(2n+1+i) - (i+j+k) \leq n$, i.e., when $1 \leq j \leq n$ and $n+1-j \leq k \leq n$. For each $i$, this counts $1 + 2 + \cdots + n = \frac{1}{2} n(n+1)$ cycle triplets. When we range over all $i$, each cycle triplet gets counted three times, so the number of cycle triplets is

$$\frac{2n+1}{3} \left( \frac{n(n+1)}{2} \right) = \frac{n(n+1)(2n+1)}{6}.$$