8. (10 points) A frog starts climbing from the bottom of a 12-meter deep well at 8:00. It climbs up 3 meters and then slides down 1 meter due to the slippery well wall. The time it takes to slide down 1 meter is one-third of the time it takes to climb up 3 meters. At 8:17, the frog reaches 3 meters below the well's mouth for the second time. The time it takes for the frog to climb from the bottom of the well to the mouth is $\qquad$ minutes.
【Analysis】First, analyze the time of "up and down" as one cycle. When the fourth cycle reaches 8 meters, it has already passed 9 meters once. The second time it reaches 9 meters is at the $\frac{1}{4}$ time of the fifth cycle. Continue analyzing accordingly.

【Solution】Solution: According to the problem:
"Up and down" time is considered as one cycle. By the fourth cycle, it reaches 8 meters, having already passed 9 meters once. The second time it reaches 9 meters is at the $\frac{1}{4}$ time of the fifth cycle.
17 minutes is $4 \frac{1}{4}$ cycles, $17 \div 4 \frac{1}{4}=4$ (minutes).
Thus, each cycle is 4 minutes, which means 1 minute per meter. By the 5th cycle, it reaches 10 meters, and then 2 more meters to the well's mouth, $4 \times 5+2$ $=22$ minutes.

Therefore, the answer is: 22