(1) Solution: Remember that students apply for jobs in companies A, B, C and D for events B, C, D and E respectively, and then P(B)=0.4, P(C)=0.5, P(D)=0.5, and P (E) = 0.6.

P(ξ=0)=0.38

P(ξ=2)=0.5

P(ξ=4)=0. 12

So the distribution list of ξ is

0 2 4

0.38 0.5 0. 12

eξ= 0×0.38+2×0.5+4×0. 12 = 1.48

(2) Solution: Since the sequence an=n2-6/5ξn+ 1(n∈N*) is strictly monotonous, the sequence 3/5 < ξ < 3/2,

Namely ξ < 5/2

P(A)= P(ξ< 5/2)= P(ξ= 0)+P(ξ= 2)= 0.88