probability question

for part b, can you use the method 1-(5/16)^3 to still get the right answer using the 1-(P'a)=p(a) formula?

YOuve not posted the working but I guess the probability (5/16)^3 represents 3 threes but doesnt condition on the sum to 9 fact.

sorry for that, so the way I approached it alternatively was considering 2,2,5 3,3,3 2,3,4 combinations which I did because the above method I stated seemed too simple for a 6 mark question, but when I usually approach 'at least' questions in probability I can usually apply the formula as I have mentioned. so why does that not apply in this case?

The
p(at least one 2) = 1 - p(no 2s)
is fine, but if you just read off the oriiginal combinations and ignore the sum to 9 constraint. The relevant combinations are
p(2,2,5)
p(3,3,3)
p(2,3,4)
(and appropriate permutations). Summing these probabilities will not give 1. Youve got to find their sum and use it in the conditional probability formula to get
p(no 2s | three dice sum to 9)
and then use
p(at least one 2 | three dice sum to 9) = 1 - p(no 2s | three dice sum to 9)