Let's solve each equation step-by-step.

### Equation 1: \((x-5)^2 = x^2 - 25\)

First, expand the left-hand side:
\[
(x-5)^2 = x^2 - 10x + 25
\]

So the equation becomes:
\[
x^2 - 10x + 25 = x^2 - 25
\]

Subtract \(x^2\) from both sides:
\[
-10x + 25 = -25
\]

Subtract 25 from both sides:
\[
-10x = -50
\]

Divide by -10:
\[
x = 5
\]

So, the solution is:
\[
x = 5
\]

### Equation 2: \(x^2 + 4x - 5 = 0\)

This is a quadratic equation. We can solve it using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = -5\).

First, calculate the discriminant:
\[
b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36
\]

Now, apply the quadratic formula:
\[
x = \frac{-4 \pm \sqrt{36}}{2 \cdot 1} = \frac{-4 \pm 6}{2}
\]

This gives us two solutions:
\[
x = \frac{-4 + 6}{2} = \frac{2}{2} = 1
\]
\[
x = \frac{-4 - 6}{2} = \frac{-10}{2} = -5
\]

So, the solutions are:
\[
x = 1 \quad \text{and} \quad x = -5
\]

### Equation 3: \((3x+1)^2 - (3x+1) = 0\)

Let \(y = 3x + 1\). Then the equation becomes:
\[
y^2 - y = 0
\]

Factor out \(y\):
\[
y(y - 1) = 0
\]

So, \(y = 0\) or \(y = 1\).

Substitute back \(3x + 1\) for \(y\):
1. \(3x + 1 = 0\)
\[
3x = -1 \implies x = -\frac{1}{3}
\]

2. \(3x + 1 = 1\)
\[
3x = 0 \implies x = 0
\]

So, the solutions are:
\[
x = -\frac{1}{3} \quad \text{and} \quad x = 0
\]

### Equation 4: \(x^2 - 3x = 2\)

Rearrange the equation to standard quadratic form:
\[
x^2 - 3x - 2 = 0
\]

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -3\), and \(c = -2\).

First, calculate the discriminant:
\[
b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot (-2) = 9 + 8 = 17
\]

Now, apply the quadratic formula:
\[
x = \frac{3 \pm \sqrt{17}}{2}
\]

So, the solutions are:
\[
x = \frac{3 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{17}}{2}
\]

### Summary of Solutions

1. \((x-5)^2 = x^2 - 25\)
\[
x = 5
\]

2. \(x^2 + 4x - 5 = 0\)
\[
x = 1 \quad \text{and} \quad x = -5
\]

3. \((3x+1)^2 - (3x+1) = 0\)
\[
x = -\frac{1}{3} \quad \text{and} \quad x = 0
\]

4. \(x^2 - 3x = 2\)
\[
x = \frac{3 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{17}}{2}
\]