(i) Given \(f(0) = 10\), we have:

\(a + b = 10\)

Given \(f\left( \frac{2}{3}\pi \right) = 1\), we have:

\(a - \frac{b}{2} = 1\)

Solving these equations:

From \(a + b = 10\), we get \(a = 10 - b\).

Substitute into \(a - \frac{b}{2} = 1\):

\(10 - b - \frac{b}{2} = 1\)

\(10 - \frac{3b}{2} = 1\)

\(\frac{3b}{2} = 9\)

\(b = 6\)

\(a = 10 - 6 = 4\)

Thus, \(a = 4\) and \(b = 6\).

(ii) The range of \(f(x) = a + b \cos x\) is from \(a - b\) to \(a + b\).

\(a - b = 4 - 6 = -2\)

\(a + b = 4 + 6 = 10\)

Therefore, the range is \(-2\) to \(10\).

(iii) To find \(f\left( \frac{5}{6}\pi \right)\):

\(\cos\left( \frac{5}{6}\pi \right) = -\cos\left( \frac{1}{6}\pi \right) = -\frac{\sqrt{3}}{2}\)

\(f\left( \frac{5}{6}\pi \right) = 4 + 6\left(-\frac{\sqrt{3}}{2}\right)\)

\(= 4 - 3\sqrt{3}\)