I'm starting to get more confident that my earlier answer to (ii) is correct as it is numerically equal to my answer using a different method.
My previous answer was 40C9 + 2(40C8) + 40C7.
This is the same as 42C9.
The logic is as follows:
The number of ways to pick 9 numbers from the first 42 integers is 42C9. If we arrange them in increasing order and add 1 to the 2nd number, 2 to the 3rd number, 3 to the 4th number, ... 8 to the 9th number, then we get a set of numbers ranging from 1 to 50 which must be separated by at least 2, which is what is required.
Following that logic, I'm guessing that if we instead required each pair of numbers to be separated by at least 3 (ie. at least two integers between them) then there would be 34C9 possibilities.
This is consistent with expectations as we increase the gap. If we require 5 numbers between any pair we get 10C9=10. If we think about this, a possible combination is 1, 7, 13, 19, 25, 31, 37, 43, 49. The only way of getting other combinations that work is to increase each of the numbers one at a time starting from the last number, giving 10 possibilities.
And if we now increase the size of the gap to 6, we get 2C9, which is meaningless as expected.