Four. (50 points) In a plane, there are 12 points, no three of which are collinear. Each pair of points is connected by a line segment (or edge). These line segments are colored in red and blue, with each line segment being colored exactly one color. Among them, the number of red line segments starting from a certain point $A$ is odd, while the number of red line segments starting from the other 11 points are all different. Find the number of triangles with the given points as vertices and all sides red, and the number of triangles with two sides red and one side blue.
Four, Solution 1 Notice that the number of red line segments drawn from each point can only be one of the values $0,1, \cdots, 11$, and $0$ and $11$ cannot appear simultaneously. Therefore, there are two possible scenarios:
(1) $0,1, \cdots, 10$;
(2) $1,2, \cdots, 11$.
If it is scenario (2), when point $A$ draws $2 k+1$ red line segments, $\frac{1}{2}\left[\frac{(1+11) \times 11}{2}+2 k+1\right]$ is not an integer. Therefore, it can only be scenario (1).

Thus, the other 11 points each draw $0,1, \cdots, 10$ red line segments.

At this point, let's assume point $A_{i}(i=1,2, \cdots, 11)$ draws $i-1$ red line segments (referred to as red edges), and point $A$ draws $2 k+1$ red edges. Then point $A_{11}$ is connected to the other 10 points $A, A_{2}, A_{3}, \cdots, A_{10}$ except point $A_{1}$ with red edges; point $A_{0}$ is connected to the other 9 points $A, A_{3}, A_{4}, \cdots, A_{10}$ except points $A_{1}$ and $A_{2}$ with red edges.

Similarly, point $A_{9}$ is connected to 8 points except points $A_{1}, A_{2}, A_{3}$; point $A_{8}$ is connected to 7 points except points $A_{1}, A_{2}, A_{3}, A_{4}$; point $A_{7}$ is connected to 6 points except points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$. Therefore, points $A_{11}, A_{10}, A_{9}, A_{8}, A_{7}$ are all connected to point $A$ with red edges.

Since point $A_{6}$ only connects 5 red edges (already connected to $A_{11}, A_{10}, A_{9}, A_{8}, A_{7}$), point $A_{6}$ does not connect to point $A$ with a red edge.

Similarly, points $A_{5}, A_{4}, A_{3}, A_{2}, A_{1}$ do not connect to point $A$ with red edges.
Thus, point $A$ connects 5 red edges.
$$
\begin{array}{l}
\text { Let } M=\left\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\right\}, \\
N=\left\{A, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}\right\} .
\end{array}
$$

Then any two points in $M$ are not connected by red edges, while any two points in $N$ are connected by red edges, and any point $A_{i}$ $(1 \leqslant i \leqslant 6)$ in $M$ is connected to $i-1$ points in $N$ with red edges. Therefore, there is no triangle with all three sides red with $A_{1}, A_{2}$ as vertices, and there are $\mathrm{C}_{i-1}^{2}$ triangles with all three sides red with $A_{i}(3 \leqslant i \leqslant 6)$ as vertices, and there are $\mathrm{C}_{6}^{3}$ triangles with all three sides red with any three points in $N$ as vertices. Thus, the number of triangles with all three sides red is
$$
\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{5}^{2}+\mathrm{C}_{6}^{3}=\mathrm{C}_{6}^{3}+\mathrm{C}_{6}^{3}=40 \text {. }
$$

Triangles with two red sides and one blue side are divided into two categories:
(i) One vertex $A_{i} \in M$, the other two vertices belong to $N$, and the two sides drawn from $M$ to $N$ are one red and one blue (the two points in $N$ are connected by red edges). The number of such triangles is
$$
\sum_{i=1}^{6} C_{i}^{1} \cdot C_{6-i}^{1}=35 \text {. }
$$
(ii) One vertex is $P \in N(P$ is $A$ or $\left.A_{j}(7 \leqslant j \leqslant 11)\right)$, the other two vertices belong to $M$, and the two sides drawn from point $P$ to $M$ are both red $(M$'s two points are connected by blue edges). The number of such triangles is
$$
\sum_{i=7}^{11} \mathrm{C}_{i-6}^{2}=20 \text {. }
$$

Thus, the number of triangles with two red sides and one blue side is $35+20=55$.
Therefore, the required numbers are 40,55.
Solution 2 From Solution 1, we know that the number of triangles with all three sides red is 40.
For any $i \geqslant 3$, the number of triangles with $A_{i}$ as a vertex and the two sides connected to $A_{i}$ both red is $\mathrm{C}_{i-1}^{2}$, and the number of triangles with $A$ as a vertex and the two sides connected to $A$ both red is $\mathrm{C}_{5}^{2}$. Adding up all these triangles, we get
$$
\sum_{i=3}^{11} \mathrm{C}_{\mathrm{i}-1}^{2}+\mathrm{C}_{5}^{3}=\mathrm{C}_{11}^{3}+\mathrm{C}_{5}^{3}=175 \text {. }
$$

In the above sum, each triangle with two red sides and one blue side is counted once, and each triangle with all three sides red is counted three times. Therefore, the number of triangles with two red sides and one blue side is
$175-3 \times 40=55$.
Thus, the required numbers are $40, 55$.
(Shen Wenxuan, Hunan Normal University Mathematical Olympiad Research Institute, 410081, Yang Mingliang, Hunan Normal University High School, 410081)