(i) To show that the curve and the line meet for all values of \(k\), equate the equations:

\(2x^2 - 3x + 1 = kx + k^2\)

Rearrange to form a quadratic equation:

\(2x^2 - (3+k)x + (1-k^2) = 0\)

For the curve and line to meet, the discriminant \(b^2 - 4ac\) must be non-negative:

\((3+k)^2 - 4 \times 2 \times (1-k^2) \geq 0\)

\(9 + 6k + k^2 - 8 + 8k^2 \geq 0\)

\(9k^2 + 6k + 1 \geq 0\)

\((3k + 1)^2 \geq 0\)

This is always true, hence the curve and line meet for all values of \(k\).

(ii) For the line to be tangent to the curve, the discriminant must be zero:

\((3+k)^2 - 4 \times 2 \times (1-k^2) = 0\)

\(9 + 6k + k^2 - 8 + 8k^2 = 0\)

\(9k^2 + 6k + 1 = 0\)

Solving for \(k\), we find \(k = -\frac{1}{3}\).

Substitute \(k = -\frac{1}{3}\) into the line equation:

\(y = -\frac{1}{3}x + \left(-\frac{1}{3}\right)^2 = -\frac{1}{3}x + \frac{1}{9}\)

Equate with the curve equation:

\(2x^2 - 3x + 1 = -\frac{1}{3}x + \frac{1}{9}\)

\(2x^2 - \frac{8}{3}x + \frac{8}{9} = 0\)

Solving gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the line equation to find \(y\):

\(y = -\frac{1}{3} \times \frac{2}{3} + \frac{1}{9} = -\frac{1}{9}\)

Thus, the point of tangency is \(\left( \frac{2}{3}, -\frac{1}{9} \right)\).