(i) Since the coefficients of the polynomial are real, the complex conjugate of any complex root must also be a root. Therefore, if -1 + \sqrt{3}i is a root, then -1 - \sqrt{3}i is also a root.

(ii) Substitute \(x = -1 + \sqrt{3}i\) into the equation \(kx^3 + 5x^2 + 10x + 4 = 0\) and expand \(x^2\) and \(x^3\) using \(i^2 = -1\).

Calculate \(x^2 = (-1 + \sqrt{3}i)^2 = 1 - 2\sqrt{3}i - 3 = -2 - 2\sqrt{3}i\).

Calculate \(x^3 = (-1 + \sqrt{3}i)(-2 - 2\sqrt{3}i) = 2 + 2\sqrt{3}i + 2\sqrt{3}i + 6i^2 = 2 + 4\sqrt{3}i - 6 = -4 + 4\sqrt{3}i\).

Substitute these into the equation:

\(k(-4 + 4\sqrt{3}i) + 5(-2 - 2\sqrt{3}i) + 10(-1 + \sqrt{3}i) + 4 = 0\).

Simplify and equate real and imaginary parts to zero to find \(k = 2\).

With \(k = 2\), the equation becomes \(2x^3 + 5x^2 + 10x + 4 = 0\).

\(Using the roots -1 + \sqrt{3}i and -1 - \sqrt{3}i, find the quadratic factor:\)

\((x + 1 - \sqrt{3}i)(x + 1 + \sqrt{3}i) = x^2 + 2x + 4\).

Divide the cubic by this quadratic to find the third root:

\(2x^3 + 5x^2 + 10x + 4 = (x^2 + 2x + 4)(2x + 1)\).

Thus, the third root is \(x = -\frac{1}{2}\).