(i) To find where \(f\) is decreasing, we need \(f'(x) < 0\).

\(3(2x-1)^{\frac{1}{2}} - 6 < 0\)

\(3(2x-1)^{\frac{1}{2}} < 6\)

\((2x-1)^{\frac{1}{2}} < 2\)

\(2x-1 < 4\)

\(2x < 5\)

\(x < \frac{5}{2}\)

Since \(x > \frac{1}{2}\) is given, the set of values is \(\frac{1}{2} < x < \frac{5}{2}\).

(ii) We integrate \(f'(x) = 3(2x-1)^{\frac{1}{2}} - 6\) to find \(f(x)\).

\(f(x) = \int (3(2x-1)^{\frac{1}{2}} - 6) \, dx\)

\(f(x) = \left[ \frac{3}{2} (2x-1)^{\frac{3}{2}} \cdot \frac{1}{2} \right] - 6x + c\)

\(f(x) = (2x-1)^{\frac{3}{2}} - 6x + c\)

Given \(f(1) = -3\), substitute \(x = 1\):

\(-3 = (2 \cdot 1 - 1)^{\frac{3}{2}} - 6 \cdot 1 + c\)

\(-3 = 1 - 6 + c\)

\(c = 2\)

Thus, \(f(x) = (2x-1)^{\frac{3}{2}} - 6x + 2\).