(i) To find the equation of the tangent at \(A\), we first find the derivative of the curve:

\(\frac{dy}{dx} = \frac{-10}{(2x+1)^2} \times 2\).

At \(A\), \(y = 0\), so \(x = 2\).

The slope \(m\) at \(x = 2\) is \(\frac{-4}{5}\).

The equation of the tangent is \(y = \frac{-4}{5}(x - 2)\).

Rearranging gives \(5y + 4x = 8\).

(ii) The point \(C\) is \((0, 1.6)\).

The distance \(AC\) is \(\sqrt{(1.6)^2 + 2^2} = 2.56\).