The first question b takes 20 from 1, if b >: 16, ξ1+ξ 2 = 20+5b > 100, where w = 0, b is less than or equal to 16 and w = ξ1+ξ 2; So the expectation of W is: ∑(20+5b)/20=50, and the sum is from 1 to16;

That is (20+5×1+20+5× 2+) ...+20+50×16)/20 = 50.

The second problem is that when a is determined, b >；; 20-a, ξ 1+ξ2=5(a+b)> 100, when w=0 and b is less than or equal to 20-a,

w =ξ 1+ξ2 = 5(a+b)；

So the expectation of W is: ∑ 5 (a+b)/20; B from 1 to 20-a;

∑5(a+b)/20 & gt； 5a

That is, (20-a) * 5a/20+(1+2+...+20-a) * 5/20 > 5a.

20a-5a * a+(2 1-a)(20-a)/2 & gt； 20a

4a * a+4 1a-420 & lt； 0

6 at most, 5a at most 30.