(i) The general term in the expansion of \(\left( x - \frac{2}{x} \right)^6\) is given by \(T_r = \binom{6}{r} x^{6-r} \left( -\frac{2}{x} \right)^r\).

For the term to be independent of \(x\), the powers of \(x\) must cancel out, i.e., \(6 - r - r = 0\), which gives \(r = 3\).

Substituting \(r = 3\) into the general term, we get:

\(T_3 = \binom{6}{3} x^{6-3} \left( -\frac{2}{x} \right)^3 = 20 \cdot x^3 \cdot \left( -\frac{8}{x^3} \right) = -160\).

Thus, the term independent of \(x\) is \(-160\).

(ii) The term independent of \(x\) in the product \(\left( 2 + \frac{3}{x^2} \right) \left( x - \frac{2}{x} \right)^6\) is found by considering the contributions from both terms.

First, find the term in \(\left( x - \frac{2}{x} \right)^6\) that results in \(x^2\):

\(T_r = \binom{6}{r} x^{6-r} \left( -\frac{2}{x} \right)^r\).

For \(x^2\), \(6 - r - r = 2\), which gives \(r = 2\).

Substituting \(r = 2\), we get:

\(T_2 = \binom{6}{2} x^{6-2} \left( -\frac{2}{x} \right)^2 = 15 \cdot x^4 \cdot \frac{4}{x^2} = 60x^2\).

Now, consider the product:

\(2 \cdot (-160) + \frac{3}{x^2} \cdot 60x^2 = -320 + 180 = -140\).

Thus, the term independent of \(x\) is \(-140\).