Answer :

There are thirty-six pairs of times when the hands exactly change places between three p.m. and midnight. 
The number of pairs of times from any hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. 
In the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, the required answer.

The first pair of times is 3 hr. 21⁵⁷/₁₄₃ min. and 4 hr. 16¹¹²/₁₄₃ min., and the last pair is 10 hr. 59⁸³/₁₄₃ min. and 11 hr. 54¹³⁸/₁₄₃ min. 
I will not give all the remainder of the thirty-six pairs of times, but supply a formula by which any of the sixty-six pairs that occur from midday to midnight may be at once found:

For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10 (where nought stands for 12 o'clock midday); and b may represent any hour, later than a, up to 11.

By the aid of this formula there is no difficulty in discovering the answer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58¹⁰⁶/₁₄₃ min. and 11 hr. 44¹²⁸/₁₄₃ min., the latter being the time when the minute hand is nearest of all to the point IX—in fact, it is only ¹⁵/₁₄₃ of a minute distant.

Readers may find it instructive to make a table of all the sixty-six pairs of times when the hands of a clock change places. 
An easy way is as follows: Make a column for the first times and a second column for the second times of the pairs. 
By making a = 0 and b = 1 in the above expressions we find the first case, and enter hr. 5⁵/₁₄₃ min. at the head of the first column, and 1 hr. 0⁶⁰/₁₄₃ min. at the head of the second column. 
Now, by successively adding 5⁵/₁₄₃ min. in the first, and 1 hr. 0⁶⁰/₁₄₃ min. in the second column, we get all the eleven pairs in which the first time is a certain number of minutes after nought, or mid-day. 

Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the ten pairs after 1 o'clock. 
Then there is another "jump," and you will be able to get by addition all the ninepairs after 2 o'clock. 
And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." 
In this way we get under the successive hours, 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees with the formula in the first paragraph of this article.

Some time ago the principal of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" 
His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. 
At length some of his readers convinced him that the answer is, "At 5⁵/₁₄₃ min. past XII;" and this he finally gave as correct, together with the reason for it that at that time the time indicated is the same whichever hand you may assume as hour hand!