Solve for x in the equations: 3.3.2 5^(x+2) = 1/25 and 3.3.1 27.3^x = 1

Here are the solutions to the equations.

3.3.1 Solve: $27^{3x} = 1$

Step 1: Express the right side of the equation as a power of 27. Any non-zero number raised to the power of 0 is 1. $27^{3x} = 27^0$

Step 2: Since the bases are the same, set the exponents equal to each other. $3x = 0$

Step 3: Solve for $x$. $x = \frac{0}{3}$ $x = 0$ The final answer is $\boxed{0}$.

3.3.2 Solve: $5^{x+2} = \frac{1}{25}$

Step 1: Express the right side of the equation as a power of 5. $25 = 5^2$ So, $\frac{1}{25} = \frac{1}{5^2}$

Step 2: Use the rule $a^{-n} = \frac{1}{a^n}$ to rewrite $\frac{1}{5^2}$. $\frac{1}{5^2} = 5^{-2}$

Step 3: Substitute this back into the original equation. $5^{x+2} = 5^{-2}$

Step 4: Since the bases are the same, set the exponents equal to each other. $x+2 = -2$

Step 5: Solve for $x$. $x = -2 - 2$ $x = -4$ The final answer is $\boxed{-4}$.