For 6a, yes, the sum should have been of, say, (j/p).

For 6b, the sum should have been of ((aj+b)/p).

For 6a, the point is that half of the j's have (j/p) = 1, and the other half have (j/p)=-1

For 6b, since a is invertible, as j ranges over 1 through p-1, the set {1, ..., p-1} is the same as the set (a*1 + b, a*2+b, ..., a*(p-1)+b).

Best,
jda