(i) The amount of salt obtained each week forms a geometric progression with the first term \(a = 8000\) kg and a common ratio \(r = 1.02\). The amount of salt obtained in the 12th week is given by the formula for the \(n\)-th term of a geometric progression:

\(a_n = a imes r^{n-1}\)

Substituting the values, we have:

\(a_{12} = 8000 imes (1.02)^{11}\)

\(a_{12} \approx 9950 \text{ kg}\)

(ii) The total amount of salt obtained in the first 12 weeks is the sum of the first 12 terms of the geometric progression. The sum \(S_n\) of the first \(n\) terms of a geometric progression is given by:

\(S_n = a \frac{r^n - 1}{r - 1}\)

Substituting the values, we have:

\(S_{12} = 8000 \frac{(1.02)^{12} - 1}{1.02 - 1}\)

\(S_{12} \approx 107000 \text{ kg}\)