Given the first term \(a = 12\) and the second term \(ar = -6\), we find the common ratio \(r\) by solving:

\(ar = -6 \Rightarrow 12r = -6 \Rightarrow r = -\frac{1}{2}\).

(i) The formula for the \(n\)-th term of a geometric progression is \(a_n = ar^{n-1}\). For the tenth term:

\(a_{10} = 12 \left(-\frac{1}{2}\right)^{9} = 12 \times -\frac{1}{512} = -\frac{3}{128}\).

(ii) The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\), provided \(|r| < 1\):

\(S_\infty = \frac{12}{1 - (-\frac{1}{2})} = \frac{12}{1 + \frac{1}{2}} = \frac{12}{\frac{3}{2}} = 8\).