12. (15 points) 10 white balls numbered from 1 to 10 are arranged in a row. Now, the balls are painted red according to the following method:
（1）Paint 2 balls;
（2）The difference in the numbers of the 2 painted balls is greater than 2. How many different painting methods are there?
【Analysis】This problem uses the enumeration method, where the number of the first ball to be colored is less than that of the second ball. Since $8+2=10$, to ensure the difference in numbers is greater than 2, the maximum number for the second ball is 7. Therefore, the first ball can be any one of the numbers from $1 \sim 7$. We will discuss each case individually and then add up the number of ways for each case.

【Solution】Solution: If the first ball is numbered 1, then the other ball can be numbered $4 \sim 10$; there are 7 different cases; if the first ball is numbered 2, then the other ball can be numbered $5 \sim 10$; there are 6 different cases; if the first ball is numbered 3, then the other ball can be numbered $6 \sim 10$; there are 5 different cases; if the first ball is numbered 4, then the other ball can be numbered $7 \sim 10$; there are 4 different cases; if the first ball is numbered 5, then the other ball can be numbered $8 \sim 10$; there are 3 different cases; if the first ball is numbered 6, then the other ball can be numbered $9 \sim 10$; there are 2 different cases; if the first ball is numbered 7, then the other ball can be numbered 10; there is 1 different case; 
Therefore, the number of different coloring methods is:
$$
7+6+5+4+3+2+1=28 \text{ (ways). }
$$

Answer: There are 28 different coloring methods.