I think there is a much simpler solution than all provided so far.

Consider the following three facts:

 - hour hand will rotate 2 full times in a day
 - minute hand will rotate 24 full times in a day
 - second hand will rotate 1440 full times in a day

So then:

(a) the minute and hour hands, will meet exactly:
>! 22 times: 24 - 2

(b) the minute and second hands, will meet exactly:
>! 1416 times: 1440 - 24

(c) the hour and second hands, will meet exactly:
>! 1438 times: 1440 - 2

(d) all three hands, will meet exactly:
>! twice: only at exactly 12:00:00 o'clock (noon and midnight)

Simply because:
>! the faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes. But one hand lapping another won't necessarily coincide with the third hand being there, so in the third case we can only measure by finding the LCD of the three periods, in this case 2.