a) The answer is $|T| = 2n(2n-1)(2n-2)$.

b) Consider the set
$$A = \{(i, j) : 1 \le i, j \le n, i \ne j\} \cup \{(i, j) : n+1 \le i, j \le 2n, i \ne j\}.$$
Clearly, $|A| = 2 \cdot A_n^2 = 2n(n-1)$. Furthermore, for all $(x, y, z) \in T$ there always exist two in three numbers $x, y, z$ belong to the same set $\{1, 2, \dots, n\}$ or $\{n+1, n+2, \dots, 2n\}$. Without loss of generality, assume that $x, y$ belong to the same set then $(x, y) \in A$, which implies $\{(x, y), (x, z), (y, z)\} \cap A \ne \emptyset$.

c) Consider the graph $G_1 = (V, E_1)$, $V$ is the set of vertices that represents for $1, 2, \dots, 2n$ and $u, v$ are connected when a set $A$ which is connected with $T$ does not have the ordered pair $(u, v)$.
From the given condition, at least one of three numbers $\{x, y, z\}$ belongs to $A$, otherwise there exists three segments which connect $(x, y)$; $(y, z)$; $(z, x)$. Without loss of generality, suppose that $x > y > z$ then $G_1$ contains $(x, y)$, $(x, z)$, $(y, z)$, which means when we consider the triple $(x, y, z)$ then none of these ordered pairs belong to $T$, contradiction. Therefore, $G_1$ does not contain a triangle.
By Mantel - Turan theorem, the number of edges in $G$ cannot exceed $\frac{(2n)^2}{4} = n^2$. Hence, the number of edges in the complementary graph is at least $\binom{2n}{2} - n^2 = n(n-1)$. Similarly, we can show that there are at least $n(n-1)$ pairs $(u, v)$ $(u < v)$ in $A$. Hence, $T$ has at least $2n(n-1)$ ordered pairs. $\Box$