Let's proceed by solving the quadratic equations using the method of completing the square. I'll begin with equation (a) and work through them step by step.

a) $2x^2 - 6x + 1 = 0$

1. Divide by 2 to simplify the equation: $x^2 - 3x + \frac{1}{2} = 0$

2. Move the constant term to the other side: $x^2 - 3x = -\frac{1}{2}$

3. Complete the square: Take half of the coefficient of $x$, square it, and add it to both sides: $\left( \frac{-3}{2} \right)^2 = \frac{9}{4}$ Add $\frac{9}{4}$ to both sides: $x^2 - 3x + \frac{9}{4} = -\frac{1}{2} + \frac{9}{4}$

4. Simplify the right-hand side: $x^2 - 3x + \frac{9}{4} = \frac{-2 + 9}{4} = \frac{7}{4}$

5. Rewrite the left-hand side as a perfect square: $\left( x - \frac{3}{2} \right)^2 = \frac{7}{4}$

6. Take the square root of both sides: $x - \frac{3}{2} = \pm \frac{\sqrt{7}}{2}$

7. Solve for $x$: $x = \frac{3}{2} \pm \frac{\sqrt{7}}{2}$

Thus, the solutions are: $x = \frac{3 + \sqrt{7}}{2} \quad \text{or} \quad x = \frac{3 - \sqrt{7}}{2}$

Would you like me to continue solving the rest of the equations, or would you like more details on any specific part?