Let's solve each equation step by step.

### a) \(\frac{4}{x} + 5 = \frac{1}{2}\)

1. Subtract 5 from both sides:
\[
\frac{4}{x} = \frac{1}{2} - 5
\]

2. Convert 5 to a fraction with a denominator of 2:
\[
\frac{1}{2} - \frac{10}{2} = \frac{1 - 10}{2} = \frac{-9}{2}
\]
So,
\[
\frac{4}{x} = \frac{-9}{2}
\]

3. Cross-multiply to solve for \(x\):
\[
4 \cdot 2 = -9 \cdot x
\]
\[
8 = -9x
\]

4. Divide by -9:
\[
x = \frac{8}{-9} = -\frac{8}{9}
\]

### b) \(\frac{2x^2}{3x - 5} - 2x = 0\)

1. Move \(2x\) to the other side:
\[
\frac{2x^2}{3x - 5} = 2x
\]

2. Cross-multiply:
\[
2x^2 = 2x(3x - 5)
\]
This simplifies to:
\[
2x^2 = 6x^2 - 10x
\]

3. Rearranging gives:
\[
0 = 6x^2 - 10x - 2x^2
\]
\[
0 = 4x^2 - 10x
\]

4. Factor out \(2x\):
\[
0 = 2x(2x - 5)
\]

5. Set each factor to zero:
\[
2x = 0 \quad \Rightarrow \quad x = 0
\]
\[
2x - 5 = 0 \quad \Rightarrow \quad 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2}
\]

### Final Solutions:
- From (a): \(x = -\frac{8}{9}\)
- From (b): \(x = 0\) and \(x = \frac{5}{2}\)