(i) Since \((x + 2)\) is a factor of \(f(x)\), substituting \(x = -2\) into \(f(x)\) should yield zero:

\(f(-2) = (-2)^3 + a(-2)^2 - a(-2) + 14 = 0\)

\(-8 + 4a + 2a + 14 = 0\)

\(6a + 6 = 0\)

\(6a = -6\)

\(a = -1\)

(ii) With \(a = -1\), the polynomial becomes:

\(f(x) = x^3 - x^2 + x + 14\)

We need to show that \(f(x) = 0\) has only one real root. First, find the quadratic factor by dividing \(f(x)\) by \((x + 2)\):

\(f(x) = (x + 2)(x^2 - 3x + 7)\)

Now, use the discriminant of the quadratic \(x^2 - 3x + 7\):

\(b^2 - 4ac = (-3)^2 - 4(1)(7) = 9 - 28 = -19\)

Since the discriminant is negative, \(x^2 - 3x + 7 = 0\) has no real roots. Therefore, \(f(x) = 0\) has only one real root, which is \(x = -2\).