Let's solve the problems step by step.

### a) \( 2 \left( \frac{1}{2} - \frac{1}{x} \right) = \frac{3}{2} - \frac{1}{4} \)

1. Simplify the right side:
\[
\frac{3}{2} - \frac{1}{4} = \frac{6}{4} - \frac{1}{4} = \frac{5}{4}
\]

2. The equation becomes:
\[
2 \left( \frac{1}{2} - \frac{1}{x} \right) = \frac{5}{4}
\]

3. Distribute the 2:
\[
2 \cdot \frac{1}{2} - \frac{2}{x} = 1 - \frac{2}{x}
\]

4. Set equal to \(\frac{5}{4}\):
\[
1 - \frac{2}{x} = \frac{5}{4}
\]

5. Solve for \(\frac{2}{x}\):
\[
1 - \frac{5}{4} = \frac{2}{x}
\]
\[
\frac{4}{4} - \frac{5}{4} = \frac{2}{x}
\]
\[
-\frac{1}{4} = \frac{2}{x}
\]

6. Solve for \(x\):
\[
x = -8
\]

### b) \( \left(\frac{1}{2} + 2x\right)(2x-3) = 0 \)

1. Set each factor to zero:
\[
\frac{1}{2} + 2x = 0 \quad \text{and} \quad 2x-3 = 0
\]

2. Solve \(\frac{1}{2} + 2x = 0\):
\[
2x = -\frac{1}{2}
\]
\[
x = -\frac{1}{4}
\]

3. Solve \(2x-3 = 0\):
\[
2x = 3
\]
\[
x = \frac{3}{2}
\]

4. Solutions for \(x\) are \(x = -\frac{1}{4}\) and \(x = \frac{3}{2}\).

### c) \( \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \frac{1}{11 \cdot 14} + \cdots + \frac{1}{x \cdot (x+3)} = \frac{101}{1540} \)

1. General term: \(\frac{1}{x(x+3)} = \frac{1}{3} \left( \frac{1}{x} - \frac{1}{x+3} \right)\)

2. The series becomes a telescoping series:
\[
\frac{1}{3} \left( \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \frac{1}{11} - \cdots - \frac{1}{x+3} \right)
\]

3. Simplifies to:
\[
\frac{1}{3} \left( \frac{1}{5} - \frac{1}{x+3} \right) = \frac{101}{1540}
\]

4. Multiply through by 3:
\[
\frac{1}{5} - \frac{1}{x+3} = \frac{303}{1540}
\]

5. Solve for \(\frac{1}{x+3}\):
\[
\frac{1}{x+3} = \frac{1}{5} - \frac{303}{1540}
\]

6. Convert to a common denominator:
\[
\frac{1}{5} = \frac{308}{1540}
\]
\[
\frac{1}{x+3} = \frac{308}{1540} - \frac{303}{1540} = \frac{5}{1540}
\]

7. Solve for \(x+3\):
\[
x+3 = \frac{1540}{5} = 308
\]
\[
x = 308 - 3 = 305
\]

Thus, \( x = -8 \) for (a), \( x = -\frac{1}{4} \) or \( x = \frac{3}{2} \) for (b), and \( x = 305 \) for (c).