Let's solve each equation step by step.

### Equation 1: \(\frac{1}{x} = \frac{5}{3(x+2)}\)

1. Cross-multiply to eliminate the fractions:
\[
1 \cdot 3(x+2) = 5 \cdot x
\]
This simplifies to:
\[
3(x + 2) = 5x
\]

2. Distribute and simplify:
\[
3x + 6 = 5x
\]

3. Move all terms involving \(x\) to one side:
\[
6 = 5x - 3x
\]
\[
6 = 2x
\]

4. Solve for \(x\):
\[
x = \frac{6}{2} = 3
\]

So, the solution for the first equation is \(x = 3\).

### Equation 2: \(\frac{x}{2x-1} = \frac{x-2}{2x+5}\)

1. Cross-multiply to eliminate the fractions:
\[
x \cdot (2x + 5) = (x - 2) \cdot (2x - 1)
\]

2. Distribute both sides:
\[
x(2x + 5) = (x - 2)(2x - 1)
\]
\[
2x^2 + 5x = 2x^2 - x - 4x + 2
\]
\[
2x^2 + 5x = 2x^2 - 5x + 2
\]

3. Move all terms to one side to set the equation to zero:
\[
2x^2 + 5x - 2x^2 + 5x - 2 = 0
\]
\[
10x - 2 = 0
\]

4. Solve for \(x\):
\[
10x = 2
\]
\[
x = \frac{2}{10} = \frac{1}{5}
\]

So, the solution for the second equation is \(x = \frac{1}{5}\).

### Equation 3: \(\frac{x^2 - 6}{x} = \frac{x + 3}{2}\)

1. Cross-multiply to eliminate the fractions:
\[
2(x^2 - 6) = x(x + 3)
\]

2. Distribute both sides:
\[
2x^2 - 12 = x^2 + 3x
\]

3. Move all terms to one side to set the equation to zero:
\[
2x^2 - 12 - x^2 - 3x = 0
\]
\[
x^2 - 3x - 12 = 0
\]

4. Factor the quadratic equation:
\[
(x - 6)(x + 2) = 0
\]

5. Solve for \(x\):
\[
x - 6 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 6 \quad \text{or} \quad x = -2
\]

So, the solutions for the third equation are \(x = 6\) and \(x = -2\).

### Summary of Solutions:
1. \(x = 3\)
2. \(x = \frac{1}{5}\)
3. \(x = 6\) or \(x = -2\)