The easier part is the rotation by two digits: This yields a valid time iff the minutes are less than $24$, which happens in $24^2=576$ cases.
If we rotate by one to the right, the ones of the minutes become the tens of the hours, so they must be between $0$ and $2$, which is the case in $18$ out of $60$ minutes. The ones of the hours become the tens of the minutes, so they must be between $0$ and $5$, which is the case in $16$ out of $24$ hours. Thus the number of valid rotations is $18\cdot16=288$.
If we rotate by one to the left, the number of valid rotations must be the same as if we rotate by one to the right, so that makes another $288$ valid rotations. Obviously if we don't rotate at all, all $24\cdot60=1440$ times are valid, so the total is $1440+576+288+288=2592$ valid rotations out of $4\cdot24\cdot60=5760$, slightly less than half.
Counting the size of a minimum generating set would be rather cumbersome by hand; at least I don't see an easy way to do it. That sort of thing is better left to our electronic friends – here's code that checks the above result and finds that a minimal generating set contains $999$ times. I wonder whether that's a coincidence...
You're exactly right, the answer is $11$ times every $12$ hours, or $1$ time every $\frac{12}{11}$ of an hour, which works out to about once every $1$ hour, $5$ minutes, and $27.27$ seconds. This isn't an average either; the amount of time that passes between each meeting of the two hands is constant.
Just think about the time 11:59, and count through $12$ hours. You'll notice that there is a meeting for each hour $12,1,2,3,\ldots,10$, but no meeting for the $11$th hour because you stop at 11:59. This makes $11$ meetings in $12$ hours.
It's not too hard to see why the time passing between two meetings is constant. Just think about the rotational symmetry.
Another (and more difficult way) to look at the problem is to calculate the angular speed of each hand. Using convenient units, we have the minute hand traveling at a rate of $60$ units per hour, and the hour hand travels at a rate of $5$ units per hour. Thus, we would like to solve the following equation:
$$60t=5t\operatorname{mod}60$$
$0$ clearly solves the equation, so we look for the smallest positive value of $t$ solving it. Thus we want to solve:
$$55t=60$$
which gives us $t=\frac{12}{11}$ hours as before.
Best Answer
Take two hands: a fast hand that completes $x$ revolutions per day, and a slow hand that completes $y$ revolutions per day. Now rotate the clock backwards, at a rate of $y$ revolutions per day: the slow hand comes to a standstill, and the fast hand slows down to $x-y$ revolutions per day. So the number of times that the hands are at right angles is $2(x-y)$.
The three hands make 2, 24, and 1440 revolutions per day, so the total is: $$2\times(24-2) + 2\times(1440-2) + 2\times(1440-24) = 5752$$