Sure, let’s go through some of the problems one by one. Here are the solutions:

### 7. \( |x-2| - |1 - 2x| = 0 \)

To solve this, consider different cases based on the expressions inside the absolute values:

**Case 1:** \( x - 2 \geq 0 \) and \( 1 - 2x \geq 0 \)
- \( x \geq 2 \) and \( x \leq \frac{1}{2} \) (no solution)

**Case 2:** \( x - 2 \geq 0 \) and \( 1 - 2x < 0 \)
- \( x \geq 2 \)
- Equation becomes: \( x - 2 - (2x - 1) = 0 \)
- Solution: \( x = 1 \) but \( x \geq 2 \) is required (no solution)

**Case 3:** \( x - 2 < 0 \) and \( 1 - 2x \geq 0 \)
- \( x < 2 \) and \( x \leq \frac{1}{2} \) \(\rightarrow x \leq \frac{1}{2} \)
- Equation becomes: \( -(x - 2) - (1 - 2x) = 0 \)
- Simplifying: \( -x + 2 - 1 + 2x = 0 \) \(\rightarrow x + 1 = 0 \) \(\rightarrow x = -1 \)
- \(-1 \leq \frac{1}{2} \), so valid solution \( x = -1\)

**Case 4:** \( x - 2 < 0 \) and \( 1 - 2x < 0 \)
- \( x < 2 \) and \( x > \frac{1}{2} \)
- Equation becomes: \( -(x - 2) - (2x - 1) = 0 \)
- Simplifying: \( -x + 2 - 2x + 1 = 0 \) \(\rightarrow -3x + 3 = 0 \) \(\rightarrow x = 1 \)
- \( \frac{1}{2} < 1 < 2 \), so valid solution \( x = 1 \)

**Final Solutions:** \( x = -1 \) and \( x = 1 \)

### 10. \( \frac{x-1}{x+2} = \frac{4}{5} \) với \( x \ne -2 \)

Cross-multiply to solve for \( x \):

\[ 5(x - 1) = 4(x + 2) \]
\[ 5x - 5 = 4x + 8 \]
\[ 5x - 4x = 8 + 5 \]
\[ x = 13 \]

### 15. \([(-0.5)^3]^{-1} = \frac{1}{64} \)

Calculate \( (-0.5)^3 \):

\[ (-0.5)^3 = -0.125 \]

Then, \((-0.125)^{-1}\) should equal \(\frac{1}{-0.125} = -8\), but it seems there might be a typographical error, as it doesn't equal \(\frac{1}{64}\).

### 16. \( 2020^{(-2)(2x-x)} = 1 \)

The expression inside simplifies:
\[ (-2)(2x-x) = -2x \]

For any \( a^0 = 1 \), we need:

\[ -2x = 0 \]
\[ x = 0 \]

Thus, \( x = 0 \).

Please let me know if you need help with more problems.