【Analysis】The 12 numbers on the clock face divide it evenly into 12 parts, with each part corresponding to a central angle of $360^{\circ} \div 12 = 30^{\circ}$, meaning the angle between each pair of adjacent numbers is $30^{\circ}$. The minute hand moves $\frac{360^{\circ}}{60} = 6^{\circ}$ per minute, and the hour hand moves $\frac{360^{\circ}}{60 \times 12} = 0.5^{\circ}$ per minute. At 8:00, the angle between the minute and hour hands is $120^{\circ}$. When Professor Wang boarded the train, the hour and minute hands on the station clock were symmetrical left and right, which is 8:00 plus $120 \div (6 + 0.5) = \frac{240}{13}$ (minutes), making it 8: $\frac{240}{13}$; at 2:15 PM, the angle between the minute and hour hands is $15 \times 6 - (60 + 0.5 \times 15) = 22.5$ (degrees). When Professor Wang got off the train, the hour and minute hands on the station clock were symmetrical top and bottom, which is 2:15 PM plus $22.5 \div (6 + 0.5) = \frac{45}{13}$ (minutes), making it 2:15 PM plus $\frac{45}{13}$ minutes = 2: $\frac{240}{13}$ PM. The difference between these two times is the duration Professor Wang spent on the train.

【Solution】At 8:00, the angle between the minute and hour hands is $120^{\circ}$, and $120 \div (6 + 0.5) = \frac{240}{13}$ (minutes). Professor Wang boarded the train at 8: $\frac{240}{13}$;
At 2:15 PM, the angle between the minute and hour hands is $15 \times 6 - (60 + 0.5 \times 15) = 22.5$ (degrees), and $22.5 \div (6 + 0.5) = \frac{45}{13}$ (minutes). Professor Wang got off the train at 2:15 PM plus $\frac{45}{13}$ minutes = 2: $\frac{240}{13}$ PM;
2: $\frac{240}{13}$ PM in 24-hour format is 14: $\frac{240}{13}$;
14: $\frac{240}{13}$ - 8: $\frac{240}{13}$ = 6 hours;
6 hours = 360 minutes.
Therefore, the answer is: 360.
4. (10 points) Professor Wang arrived at the station at 8:00 AM. When he boarded the train, the hour and minute hands of the clock on the platform were exactly symmetrical left and right. The train departed at 8:35 AM and arrived at the terminal station at 2:15 PM. When Professor Wang got off the train, the hour and minute hands of the clock on the platform were exactly symmetrical top and bottom, and he walked out of the station at exactly 3:00 PM. Therefore, the total time Professor Wang spent on the train is $\qquad$ minutes.