Shit, x? +(a-8)x-8a≤0, so (x+a)(x-8)≤0.

When a< is at -8, q = {x | 8≤x ≤- a }；; When a=-8, q = {x | x = 8 }；; When a> is at -8, q = {x |-a ≤ x ≤ 8};

Draw the situation on the number axis? Set p, q? (pictured)

(1) If a=-4 (in fact, you can take any value of -3≤a≤5), then q = {x | 4 ≤ x ≤ 8}; ∴p∩q={x|5<； X≤8}, but it is defined by p ∩ q = {x | 5.

(2) If p ∩ q = {x | 5

It is easy to know that when p ∩ q = {x | 5

(3) As can be seen from the figure, when -5≤a≤3 and Q={x|-a≤x≤8}, it can be deduced that if p ∩ q = {x | 5.

So "-5≤a≤3" is p ∩ q = {x | 5.