Answer:
The three rules that we have are:



Then let's solve the problems:
A) 2^4 = 2^(2*2) = (2^2)^2 = 4^2 = 16.
B) 3^0 = 3^(a - a) = 3^a/3^a = 1. (for any real number a)
From this we can conclude that the power 0 of any number is equal to 1.
C) 4^6*4^(-3) = 4^(6 - 3) = 4^3 = 64
D) (2^3)^2 = 2^(3*2) = 2^6 = 64