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. 

With the special case:

>! One hand lapping another won't necessarily coincide with the third hand being there.