4^(x+2)+64=0........20point!!!

I think your question have some typing error.
4^(x+2)+64=0 have no solution for x , it is because
4^k is always > 0 for any real number k , so RHS = 4^k + 64 is always > 0
= LHS.
You can see the problem by taking log :
4^(x+2)+64=0
4^(x+2) = - 64
(x+2)log4 = log (-64)
But log (-64) is not a real number!!
negative number can't be taken log !!

If the question is : 4^(x+2) - 64 =0 , then
4^(x+2) - 64 =0
4^(x+2) = 64
4^(x+2) = 4^3
x+2 = 3
x = 1

2^2(x+2)+2^6=0

2(x+2)+6=0 [Remove Base 2]

2x+4+6=0

2x+10=0

x=-5

Check [Substitue x=-5 in equation]:

4^(-5+2)+64=0

4^-3+4^3=0

0=0 ......OK