Let's solve the given equations step by step.

### Equation A: \(\frac{X^2 + 2x + 1}{x + 1} = 0\)

For a fraction to be equal to zero, the numerator must be zero (assuming the denominator is not zero).

So, we need to solve:
\[ X^2 + 2x + 1 = 0 \]

This is a quadratic equation. Let's factorize it:
\[ X^2 + 2x + 1 = (X + 1)^2 \]

So, we have:
\[ (X + 1)^2 = 0 \]

Taking the square root of both sides:
\[ X + 1 = 0 \]

Therefore:
\[ X = -1 \]

However, we must check the denominator to ensure it is not zero:
\[ x + 1 \neq 0 \]

Since \( x = -1 \) makes the denominator zero, there is no solution to this equation because it would make the original expression undefined.

### Equation B: \(\frac{4}{x - 2} - 2 = 0\)

First, let's isolate the fraction:
\[ \frac{4}{x - 2} = 2 \]

Now, solve for \( x \) by multiplying both sides by \( x - 2 \):
\[ 4 = 2(x - 2) \]

Simplify the right-hand side:
\[ 4 = 2x - 4 \]

Add 4 to both sides:
\[ 8 = 2x \]

Divide both sides by 2:
\[ x = 4 \]

So, the solution to the second equation is:
\[ x = 4 \]

### Summary:
- Equation A has no solution because \( x = -1 \) makes the denominator zero.
- Equation B has a solution: \( x = 4 \).