a) Not sure what your working means. There are not 15 combinations.
How many ways are there to choose 2 new batteries out of 4? How many ways are there to choose 1 old battery out of 2? Hence how many ways the first 3 batteries tested comprise 2 new and 1 old battery? Divide by number of ways of choosing 3 batteries of out 6.
Or What is the probability the first is new, the second is new, and last is old? How many ways can you get new, new, old batteries? Hence what is the probability
b) This is 1 - P(all are new)