In the solution to question 3 you wrote :
"writing Wp = xyz with y = a^k, 1=<k<=p, we can choose i = p!/k and then xyz = a^(p!+p)b^(p!+p), not in L"
as far as i understand , the equation should be (i-1)k = p! because to acheive a^p we need to pump y just "i-1" times
meaning (i-1)k = p! => i-1 = p!/k => i = p!/k +1
otherwise, the pumping can keep xyz in L…