To solve the equations provided in your image, we'll tackle both of them step by step.

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

1. **Isolate the squared term**:
\[
(3x - 1)^2 = 1
\]

2. **Take the square root of both sides**:
\[
3x - 1 = \pm 1
\]

3. **Solve for \(x\)**:
- When \(3x - 1 = 1\):
\[
3x = 2 \implies x = \frac{2}{3}
\]
- When \(3x - 1 = -1\):
\[
3x = 0 \implies x = 0
\]

**Solutions for Equation 1**: \(x = 0\) and \(x = \frac{2}{3}\)

---

### Equation 2: \(81x^2 - 18x - 3 = 0\)

1. **Use the quadratic formula**:
The formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 81\), \(b = -18\), and \(c = -3\).

2. **Calculate \(b^2 - 4ac\)**:
\[
b^2 - 4ac = (-18)^2 - 4 \cdot 81 \cdot (-3) = 324 + 972 = 1296
\]

3. **Find the square root**:
\[
\sqrt{1296} = 36
\]

4. **Plug into the formula**:
\[
x = \frac{18 \pm 36}{162}
\]
- When taking \(+\):
\[
x = \frac{54}{162} = \frac{1}{3}
\]
- When taking \(-\):
\[
x = \frac{-18}{162} = -\frac{1}{9}
\]

**Solutions for Equation 2**: \(x = \frac{1}{3}\) and \(x = -\frac{1}{9}\)

---

### Final Solutions
- Equation 1: \(x = 0\) and \(x = \frac{2}{3}\)
- Equation 2: \(x = \frac{1}{3}\) and \(x = -\frac{1}{9}\)