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?

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 $180^{\circ}\equiv -180^{\circ}{\pmod {360^{\circ}}}$
