Solution 1. First, consider the case where adjacent ends are different in color except for E and F. At this time,

There are four kinds of painting methods for A, three kinds of painting methods for point B adjacent to A, three kinds of painting methods for D and two kinds of painting methods for E,

At this point, there are two ways to draw C and two ways to draw F, so * * * has it.

4*3*3*2*2*2=288 species.

However, it is possible that E and F are the same color. When B and D are the same color, A and C are different colors, and E and F are the same color. At this time, there are four kinds of E and F with the same color, while for point E and point A, there are 3 * * * 2 = 6 kinds of D, and there is only one drawing from symmetric B and C. 。

So * * * has 4*3*2*=24 (species).

Therefore, there are 288-24=264 paintings that meet the requirements of the topic.