(i) We have the equations:

\(a + \frac{1}{2}b = 5\)

\(a - b = 11\)

Solving these simultaneously:

From \(a - b = 11\), we get \(a = 11 + b\).

Substitute into \(a + \frac{1}{2}b = 5\):

\(11 + b + \frac{1}{2}b = 5\)

\(11 + \frac{3}{2}b = 5\)

\(\frac{3}{2}b = -6\)

\(b = -4\)

Substitute \(b = -4\) into \(a = 11 + b\):

\(a = 11 - 4 = 7\)

Thus, \(a = 7\) and \(b = -4\).

(ii) The function \(f(x) = a + b \cos x\) has a range determined by the maximum and minimum values of \(\cos x\), which are 1 and -1, respectively.

Thus, the range of \(f(x)\) is:

\(a + b \leq f(x) \leq a - b\)

Substitute \(a = 7\) and \(b = -4\):

\(7 - 4 \leq f(x) \leq 7 + 4\)

\(3 \leq f(x) \leq 11\)

For \(f(x) = k\) to have no solution, \(k\) must be outside this range:

\(k < 3\) or \(k > 11\).