(i) Consider the equation \(2|x - 1| = 3|x|\). This can be split into two cases:

Case 1: \(2(x - 1) = 3x\)

\(2x - 2 = 3x\)

\(-2 = x\)

Case 2: \(2(1 - x) = 3x\)

\(2 - 2x = 3x\)

\(2 = 5x\)

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

Thus, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).

(ii) For the equation \(2|5^x - 1| = 3|5^x|\), we use the same approach:

Case 1: \(2(5^x - 1) = 3(5^x)\)

\(2 \cdot 5^x - 2 = 3 \cdot 5^x\)

\(-2 = 5^x\)

Case 2: \(2(1 - 5^x) = 3(5^x)\)

\(2 - 2 \cdot 5^x = 3 \cdot 5^x\)

\(2 = 5 \cdot 5^x\)

\(5^x = \frac{2}{5}\)

Taking logarithms, \(x = \log_5\left(\frac{2}{5}\right)\)

Using a calculator, \(x \approx 0.569\) to 3 significant figures.