The time is 0:00.00. Hands are moving continuously. What is the next angle when they are meet again? Answer for the 2nd question is 22 I think. And for the 1st question I think like this, but is cant solve.
HT = 30T, MT = 360T, ST = 360T × 60 = 21600T
MT - HT = 360 × n, ST - HT = 360 × m
360T - 30T = 330T = 360n 21600T - 30T = 21570T = 360m
For simplicity sake, lets figure it out for $12$ rather than $24$ hour period.
If $m$ is the number of minutes past noon, then the hour hand is at $360^\circ(\frac m{12*60})$ and the minute hand is at $360^\circ(\frac m{60}- \lfloor \frac m{60} \rfloor)$ where $0 \le m < 12*60$.
So $\frac {m}{12*60} = \frac m{60}- \lfloor \frac m{60} \rfloor$
$\lfloor \frac m{60} \rfloor = \frac m{60}(1 - \frac 1{12})= \frac {11m}{12*60}$
For $0 \le m < 12*60$ there $12$ possible values for $\lfloor \frac m{60} \rfloor$. They are $k = 0... 11$.
So $k = \frac {11m}{12*60}$
$m = \frac {12*60}{11}k$.
however note that for $k = 11$ then $m = 12*60$.
So the hands match up every $65 \frac 5{11} $ minutes and this occurs $12$ times is a $12$ hour period IF you consider that this occurs at $12$ noon and $12$ midnight. In a $24$ hour period it will occur $23$ times. Noon, midnight, and noon again, and other times each twelve hour period.
The angles will be $360^\circ(\frac m{12*60}) = \frac 12^\circ m = \frac 12^\circ\frac {12*60}{11}k = 360^\circ \frac {k}{11}$.