(i) To find the intersection points, set the equations equal:

\(\frac{1}{x} + c = cx - 3\)

Multiply through by \(x\) to clear the fraction:

\(1 + cx = cx^2 - 3x\)

Rearrange to form a quadratic equation:

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

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

Use the quadratic formula \(b^2 - 4ac\) to find the discriminant:

\((c + 3)^2 - 4c = c^2 + 6c + 9 - 4c = c^2 + 2c + 9\)

For real solutions, the discriminant must be non-negative:

\(c^2 + 2c + 9 \geq 0\)

Critical values are \(c = -1\) and \(c = -9\). Thus, \(c \leq -9\) or \(c \geq -1\).

(ii) For tangency, the discriminant must be zero. Substitute \(c = -1\) and \(c = -9\) into the quadratic equation:

For \(c = -1\):

\(-x^2 - 2x - 1 = 0\)

Solving gives \(x = -1\).

For \(c = -9\):

\(-9x^2 + 6x - 1 = 0\)

Solving gives \(x = \frac{1}{3}\).