Answer. At 1 hour $8^{2} /{ }_{11}$ min or at 1 hour $40^{10} / 11$ min. Solution. Let it be $x$ minutes past two when Petya looked at the clock. Since the minute hand moves $6^{\circ}$ per minute, and the hour hand $-0,5^{\circ}$, the hour hand at this moment formed an angle of $30^{\circ}+0,5 x^{\circ}$ with the 12 o'clock direction, and the minute hand - an angle of $6 x^{\circ}$. In one hour, the minute hand made a full revolution and returned to its original position, while the hour hand turned $30^{\circ}$. Clearly, the minute hand will be directed along the line bisecting the angle between the two positions of the hour hand. Thus, $6 x^{\circ}=\left(\left(30^{\circ}+0,5 x^{\circ}\right)+\left(60^{\circ}+0,5 x^{\circ}\right)\right) / 2$, if the minute hand lies inside the angle formed by the two positions of the hour hand, or $6 x^{\circ}-180^{\circ}=\left(\left(30^{\circ}+0,5 x^{\circ}\right)+\left(60^{\circ}+0,5 x^{\circ}\right)\right) / 2$, if not. Solving these two equations, we get the two answers given above.
4. Before starting to solve the problem, Kolya looked at the clock. It was two o'clock. Spending exactly an hour on solving it, Kolya looked at the clock again and noticed that the angle between the hour and minute hands remained the same. When did Kolya start solving the problem?