Working with Clocks I have frequently seen problems like how many times between _ and _ will the minute and hour hand be together, or be 90 degrees apart.
So if someone can give me a complete solution to the following three parts I would be grateful!
a) How would we find the number of times the minute and hour hand are together from 12:00 a.m and 12:00 p.m?
b) How many times will the minute and hour hand be diametrically opposite?
c) How many times will the minute and hour hand be 90 degrees apart?
 A: Hint: imagine placing the clock on a turntable that rotates very slowly, so that the hour hand doesn't move at all.
The clock itself rotates once counter-clockwise every 12 hours. The hour hand doesn't move at all. How many times around does the minute hand go every 12 hours?
A: The Hour hand, rotates  $360^{\circ}$ in in 12 hours $=12*60$ minutes,
in $t $ minutes rotates $\frac{t}{2}^{\circ}$ 
The Minute hand, rotates $360^{\circ}$  in $1$ hours=$60$ minutes,
in $t $ minutes rotates $6t^{\circ}$ 
So, in $t $ minutes, difference of angles between the hands is $6t^{\circ}-\frac{t}{2}^{\circ}$ $=\frac{11t}{2}^{\circ}$ 
If they make angle $\theta$ (in $^{\circ}$) between them,
  $\frac{11t}{2}^{\circ}=n360^{\circ}+\theta$,  where $n$ is any integer,
 or $t=\frac{n720^{\circ}+2\theta}{11}$
So, the minimum interval of making angle $\theta$ between them is 
$$\frac{(m+1)720^{\circ}+2\theta}{11}-\frac{m720^{\circ}+2\theta}{11} minute$$ 
$=\frac{720}{11} minute$ (Putting $n=m+1$ and $m$)
So, in  $12$  hours, they will make angle $\theta$ between them 
$$\frac{12 hours}{\frac{720}{11} minutes}$$
$=11$ times
(1)For coincidence, $\theta=0$
So, in  $12$  hours, they will coincide $=11$ times.
(2)For  diametrically opposite, $\theta=180^{\circ}$
So, in $12$ hours, they will be diametrically opposite $=11$ times.
(3) To be perpendicular,   $\theta=±90^{\circ}$ 
Clearly, for the '+' sign, there will be 11 occurrences of perpendicularity and so for the '-' sign. 
So, in $12$ hours, they will be   perpendicular $11+11=22$ times.
Observe that for $180^{\circ}$, we  don't consider '±' as $180^{\circ}\equiv -180^{\circ}{\pmod {360^{\circ}}}$
A: For all one-hour periods, the minute hand makes a full revolution. For most one-hour periods, the slow-moving hour hand will be swept over by the minute hand, with one moment of alignment. The exception to this is if the one-hour period begins with the minute hand strictly less than one-twelfth of a revolution ahead of the hour hand. The minute hand now has so far to go that it cannot catch up to the hour hand within an hour. If the minute hand starts exactly one-twelfth of a revolution ahead, then it will catch up to the hour hand exactly at the end of that hour. Of course if they start at the same place at the beginning of the one-hour period, that counts as an alignment.
Considering midnight to noon as 12 consecutive one-hour periods, where the minute hand is always starting at the 12-position and the hour hand is always starting spot on an hour, then no one of these twelve hour periods meets the criteria for the exception to a cross-over. There will be twelve points in time where the two hands are in the same direction. This counts both midnight and noon as two of those twelve times. And if you ask the same question over the time interval [midnight, noon), then you have the 11 that is in @lab's answer.
The thinking behind this answer changes if


*

*you have a $12$ hour period that starts at a time that is not a perfect clock hour, since then there will be one hour out of the 12 where the criterion for no crossover occurs

*you have a $12$ hour period that starts at a perfect clock hour other than 12 o'clock, since then one crossing would count twice: at the very end of the 11:00 hour and the very beginning of the 12:00 hour.

*your time period starts at 12 o'clock, but is more than 12 hours, since then one crossing would count twice: at the very end of the 11:00 hour and the very beginning of the 12:00 hour.


So you see it's complicated, and your example where the clock starts at 12:00 is very special.
The question about being diametrically opposite can be answered just as handily if you throw in a ghost minute hand that is opposite to the true minute hand. Now the ghost hand starts at the 6 o'clock position, and will be there at the beginning of every hour.  For each of the 12 closed one-hour intervals, there is one moment of alignment. But the 5-6 hour and 6-7 hour alignments are repeats. So there will be 11 alignments. (For the time period of [6am, 6pm], there would still be 12 alignments.)
And this thinking can be adopted for the 90 degree question too, if you first bring in a ghost hand 90 degree clockwise from the true minute hand, and then one 90 degrees counter-clockwise. Just as with diametrically opposite, there will be 11 alignments for each of the two 90 degree rotations, making a total of 22 alignments. (Although if the interval were [3am,3pm] or [6am,6pm], we would have a little more to think about.)
A: General theory
The hour hand covers 30 degrees in one hour and the minute hand covers 360 degrees.
so, effectively, minute hand covers 330 degrees more than hour hand in one hour.
To meet the hour hand or to coincide with the hour hand it has to cover 360 degrees, which will take slightly more than one hour
360/330 * 60 min = 65 5/11 minutes
(a) In 12 hours it happens for  12*60/65 5/11 minutes= 11 times.
It happens at 12.00 then 12.00 + 65 5/11 = around 1.05.
Similarly at around 2.10,3.15, 4.20, 5.25, 6.30, 7.35, 8.40, 9.45, 10.50
The next one is again at 12.00
(b)Similar to the previous one
It happens at 6.00,at around 7.05,8.10,9.15,10.20,11.25,12.30,1.35,2.40,3.45,4.50
and again at 6.00
The above two happens 11 times in 12 hours and once in every hour.
whereas 90 degree happens twice in every hour
(c) so 90 degrees happens 22 times in twelve hours 
