Assume that the errors in different pages are independent to each other.
Then, for (a), no errors on the first two pages =
For (b), exactly one error in the first three pages = 3.P[exactly 1 error in 1 page and no error in other 2 pages] (3 is multiplied because you can choose 1 page (which contains error) form 3 pages in 3 ways)
For (c), first error on the third page=
And for (d), no error on exactly two of the first five pages=
It has no meaning since Poisson distribution takes values 1,2,3,.... and so on (not 0).P(X< or = 0)......
Also, always stick to the original parameter (here 0.2) while calculating the probabilities.