8.2. Petya came home from school today at $16:45$, looked at the clock and wondered: after what time will the hands of the clock be in the same position for the seventh time since he came home from school?
Answer: 435 minutes.

Solution. The speed of the minute hand is 12 divisions/hour (one division here refers to the distance between adjacent numbers on the clock face), and the hour hand is 1 division/hour. Before the seventh meeting of the minute and hour hands, the minute hand must first "lap" the hour hand 6 times, meaning it must complete 6 full circles of 12 divisions. Let's say after this, the hour hand travels $L$ divisions until the seventh meeting. The total distance traveled by the minute hand then consists of the 72 divisions found, plus the initial 7.75 divisions separating them (since the clock shows 4:45), and the final $L$ divisions. Equate the time of movement of the hour and minute hands:

$$
\frac{L}{1}=\frac{L+72+7.75}{12} \Leftrightarrow 12 L=L+79.75 \Leftrightarrow L=7.25
$$

The hour hand will travel 7.25 divisions, which corresponds to 7.25 hours, or 435 minutes.