(1 - √2cosx)*(1 + 4tgx) = 0

We can solve this equation by setting each factor equal to zero:

1 - √2cosx = 0
√2cosx = 1
cosx = 1/√2
x = π/4 + 2πn, where n is an integer

1 + 4tgx = 0
4tgx = -1
tgx = -1/4
x = arctan(-1/4) + πn

0.5^x = 64

Rewrite 64 as 2^6:

(1/2)^x = 2^6
2^(-x) = 2^6
-x = 6
x = -6

4^x - 6*2^x + 8 = 0

Let's substitute 2^x = y:

y^2 - 6y + 8 = 0
(y - 4)(y - 2) = 0
y = 4 or y = 2

If y = 4:
2^x = 4
x = 2

If y = 2:
2^x = 2
x = 1

Therefore, the solutions for the given equations are: x = π/4 + 2πn, x = arctan(-1/4) + πn, x = -6, and x = 1,2.