(1) Accumulate the products of each sum frequency to get the sum of the 15 passengers, and divide it by 15 to get the average waiting time of the 15 passengers;

(2) According to the frequency of 15 passengers whose waiting time is less than 15 minutes, and the total is 8, the number of these 60 passengers whose waiting time is less than 10 minutes can be estimated;

(3) Number the two groups of passengers, and then list all basic events and the number of basic events of two people drawn from different groups. You can get the answer by substituting into the classical probability formula. Solution: (1)115 (2.5× 2+7.5× 6+12.5× 4+17.5× 2+22.5×1.

(2) The probability that the waiting time is less than 10 minute is (3+6)/ 15=8/ 15,

Therefore, the number of people waiting less than 10 minutes is 60×8/ 15=32.

(3) Number the third group of passengers as a 1, a2, a3, a4, and number the fourth group of passengers as b 1, B2.

Two people out of six have the following 15 basic events:

(a 1，a2)，(a 1，a3)，(a 1，a4)，(a 1，b 1)，(a 1，b2)，

(a2，a3)，(a2，a4)，(a2，b 1)，(a2，b2)，(a3，a4)，

(a3，b 1)，(a3，b2)，(a4，b 1)，(a4，b2)，(b 1，b2)，

Two of them just come from different groups and contain 8 basic events, so the probability is 8/ 15.