Let's solve each part (a, b, c, and d) step-by-step.

### a) Solve the equation \( 25 - 3(6 - x) = 22 \)

1. Distribute \( -3 \):
\[
25 - 18 + 3x = 22
\]
Simplifying:
\[
7 + 3x = 22
\]

2. Subtract 7 from both sides:
\[
3x = 15
\]

3. Divide by 3:
\[
x = 5
\]

### b) Solve the equation \( \left(\frac{2x - 11}{3} + 1\right) \cdot 0.5 = 20 \)

1. Multiply both sides by 2 to eliminate the 0.5:
\[
\frac{2x - 11}{3} + 1 = 40
\]

2. Subtract 1 from both sides:
\[
\frac{2x - 11}{3} = 39
\]

3. Multiply by 3:
\[
2x - 11 = 117
\]

4. Add 11 to both sides:
\[
2x = 128
\]

5. Divide by 2:
\[
x = 64
\]

### c) Solve the equation \( \frac{(25 - 2x)^3}{5} - 3^2 = 4^2 \)

1. Calculate \( 3^2 \) and \( 4^2 \):
\[
3^2 = 9 \quad \text{and} \quad 4^2 = 16
\]

2. Rewrite the equation:
\[
\frac{(25 - 2x)^3}{5} - 9 = 16
\]

3. Add 9 to both sides:
\[
\frac{(25 - 2x)^3}{5} = 25
\]

4. Multiply by 5:
\[
(25 - 2x)^3 = 125
\]

5. Take the cube root of both sides:
\[
25 - 2x = 5
\]

6. Solve for \( x \):
\[
2x = 20 \quad \Rightarrow \quad x = 10
\]

### d) Solve the equation \( 2 \cdot 3^x = 10 \cdot 3^{12} + 8 \cdot 27^4 \)

1. Recognize that \( 27 = 3^3 \), so \( 27^4 = (3^3)^4 = 3^{12} \):
\[
8 \cdot 27^4 = 8 \cdot 3^{12}
\]

2. Rewrite the original equation:
\[
2 \cdot 3^x = 10 \cdot 3^{12} + 8 \cdot 3^{12}
\]
Simplifying the right side:
\[
2 \cdot 3^x = (10 + 8) \cdot 3^{12} = 18 \cdot 3^{12}
\]

3. Divide both sides by 2:
\[
3^x = 9 \cdot 3^{12}
\]
(since \( 18/2 = 9 \))

4. Express 9 as \( 3^2 \):
\[
3^x = 3^2 \cdot 3^{12} = 3^{2 + 12} = 3^{14}
\]

5. Hence:
\[
x = 14
\]

### Summary of solutions:
- a) \( x = 5 \)
- b) \( x = 64 \)
- c) \( x = 10 \)
- d) \( x = 14 \)