Poisson

(a) ${e^{-4}4^6\over 6!}$ .

(b) $P(X\ge 10) =1-P(X\le 9)$ where $\lambda =8$ since this is a two week period and you expect twice as many parts used.

So, $P(X\ge 10) =1-\sum_{k=0}^9 {e^{-8}8^k\over k!}$ .

(c) Here you want three 'successes' in three weeks (binomial) where p is from (a)... $\biggl({e^{-4}4^6\over 6!}\biggr)^3$ .

Sorry, I do not understand (d).