(i) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1 - r}\), where \(a\) is the first term. Here, \(a = r^2 - 3r + 2\).

Thus, \(S = \frac{r^2 - 3r + 2}{1 - r}\).

Factor the numerator: \(r^2 - 3r + 2 = (r-1)(r-2)\).

So, \(S = \frac{(r-1)(r-2)}{1-r} = \frac{-(1-r)(r-2)}{1-r} = 2 - r\).

(ii) For the sum to infinity to exist, \(|r| < 1\). Therefore, \(1 < 2 - r < 3\).

This simplifies to \(1 < S < 3\) or \((1, 3)\).