To solve the given equations for \( x \), let's address each equation one by one.

### Equation 1: \( 8^x - 5 = 81 \)

1. Add 5 to both sides to isolate the exponential term:
\[
8^x = 86
\]
2. Take the logarithm of both sides to solve for \( x \):
\[
\log(8^x) = \log(86)
\]
3. Use the power rule of logarithms (\( \log(a^b) = b \log(a) \)):
\[
x \log(8) = \log(86)
\]
4. Solve for \( x \):
\[
x = \frac{\log(86)}{\log(8)}
\]

### Equation 2: \( 5^{2x - 1} = 12 \)

1. Take the logarithm of both sides:
\[
\log(5^{2x - 1}) = \log(12)
\]
2. Use the power rule of logarithms:
\[
(2x - 1) \log(5) = \log(12)
\]
3. Solve for \( 2x - 1 \):
\[
2x - 1 = \frac{\log(12)}{\log(5)}
\]
4. Add 1 to both sides:
\[
2x = \frac{\log(12)}{\log(5)} + 1
\]
5. Divide by 2 to solve for \( x \):
\[
x = \frac{1}{2} \left( \frac{\log(12)}{\log(5)} + 1 \right)
\]

### Equation 3: \( 2^{2x + 1} = 8 \)

1. Recognize that \( 8 \) can be written as \( 2^3 \):
\[
2^{2x + 1} = 2^3
\]
2. Since the bases are the same, set the exponents equal to each other:
\[
2x + 1 = 3
\]
3. Solve for \( x \):
\[
2x = 2 \implies x = 1
\]

### Summary of Solutions

1. For \( 8^x - 5 = 81 \):
\[
x = \frac{\log(86)}{\log(8)}
\]

2. For \( 5^{2x - 1} = 12 \):
\[
x = \frac{1}{2} \left( \frac{\log(12)}{\log(5)} + 1 \right)
\]

3. For \( 2^{2x + 1} = 8 \):
\[
x = 1
\]