So m =100-(3+3+37+15) = 42, …(2 points)

The frequency of the fourth group is n = 42 100 = 0.42,

N = 0.03+0.03+0.37+0.42+0.15 =1... (3 points)

The obtained histogram of frequency distribution is as shown on the right ... (5 points)

(2) according to the meaning of the question, more than 90 points are in the fifth and sixth groups respectively.

The sum of their frequencies is 0.42+0. 15=0.57,

The estimated number of students with 90 points or above in the whole region is 0.57×600=342...(7 points).

(III) Let three people whose test scores are within (0,30) be A, B and C respectively;

The three people whose test scores are within (30, 60) are A, B, C,

Of the 6 people who never scored more than 60, the results of randomly selecting 2 people are as follows:

(A，B)，(A，C)，(A，A)，(A，B)，(A，C)，

(B，C)，(B，a)，(B，B)，(B，C)，(C，a)，

(C，b)，(C，C)，(a，b)，(a，C)，(b，C)

* * * Yes 15...( 10 score)

Let the event where two people score no more than 30 points be event D.

Event d contains three results: (a, b), (a, c), (b, c)? ...( 1 1)

∴ The probability that the selected two people score less than 30 points is p (d) = 315 =15 ... (13 points).