To solve the equations provided, we'll tackle them one at a time.

### a) \( 5x^2 - 4x - 1 = 0 \)

This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 5 \)
- \( b = -4 \)
- \( c = -1 \)

We can use the quadratic formula to find the roots:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Plugging in the values:

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

2. Calculate the two potential solutions:
\[
x = \frac{4 \pm \sqrt{36}}{2 \cdot 5} = \frac{4 \pm 6}{10}
\]
This gives us two cases:
\[
x_1 = \frac{4 + 6}{10} = \frac{10}{10} = 1
\]
\[
x_2 = \frac{4 - 6}{10} = \frac{-2}{10} = -\frac{1}{5}
\]

Thus, the solutions to \( 5x^2 - 4x - 1 = 0 \) are:
\[
x = 1 \quad \text{and} \quad x = -\frac{1}{5}
\]

### b) \( x^3 - 3x + 2 = 0 \)

This is a cubic equation. We can look for rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term 2 over the factors of the leading coefficient (1), resulting in the possible roots \( \pm 1, \pm 2 \).

Testing these possible roots:

1. **Test \( x = 1 \)**:
\[
1^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \quad \text{(works)}
\]

Thus, \( x = 1 \) is a root.

2. **Perform synthetic division of \( x^3 - 3x + 2 \) by \( x - 1 \)**:
- Coefficients: \( 1 \) (for \( x^3 \)), \( 0 \) (for \( x^2 \)), \( -3 \) (for \( x \)), \( 2 \) (constant)

```
1 | 1 0 -3 2
| 1 1 -2
---------------------
1 1 -2 0
```

This gives us:
\[
x^3 - 3x + 2 = (x - 1)(x^2 + x - 2)
\]

We can factor \( x^2 + x - 2 \):
\[
x^2 + x - 2 = (x - 1)(x + 2)
\]

Thus, the full factorization becomes:
\[
x^3 - 3x + 2 = (x - 1)^2 (x + 2)
\]

Now we can find the roots:
- From \( (x - 1)^2 = 0 \): \( x = 1 \) (double root)
- From \( x + 2 = 0 \): \( x = -2 \)

The solutions to \( x^3 - 3x + 2 = 0 \) are:
\[
x = 1 \quad (double) \quad \text{and} \quad x = -2
\]

### Summary of Solutions
- For \( 5x^2 - 4x - 1 = 0 \): \( x = 1, -\frac{1}{5} \)
- For \( x^3 - 3x + 2 = 0 \): \( x = 1 \) (double root), \( x = -2 \)