(Poisson distribution) I don't understand this question and i need help please?

Strictly speaking this should all be done by the Binomial distribution. Remember the conditions for a binomial distribution are
a) fixed number (n) of trials each with only two outcomes
b) constant probability of success (p)
c) independence of results

These all apply to both the testing of an individual batch and the batches taken as a group. However, the Poisson distribution is used as an approximation to the Binomial if n is large, p is small and you are only interested in a small number of successes. If you do use the Poisson distribution for the first part then the calculation goes like this.

For an individual batch mean = m = 40*0.04 = 1.6
P(x >= 4) = 1 - P(x = 0 or 1 or 2 or 3)
= 1 - (1 + 1.6 + (1/2)*1.6^2 + (1/6)*1.6^3)*e^(-1.6) = 0.07881 (5 dec. pl.)
[ N.B. Binomial answer to this part would be 0.07484 (5 dec. pl.) ]

For the batches as a group I think you must use the Binomial distribution because n is not large enough to justify the use of Poisson.
It will have p = 0.07881, q = 0.92119, n = 10
P(x >= 2) = 1 - P(x = 0 or 1)
= 1 - 0.92119^10 - 10*(0.92119^9)*0.07881 = 0.1835

This is not that close to the answer when the Binomial distribution is used for both parts 0.1690 (4 dec. pl.) because in the first part the conditions for using Poisson were dubious.