(i) To find the first three terms in the expansion of \(\left( x - \frac{2}{x} \right)^6\), we use the binomial theorem:

\(\left( x - \frac{2}{x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left( -\frac{2}{x} \right)^k\)

Calculating the first three terms:

For \(k = 0\): \(\binom{6}{0} x^6 = x^6\)

For \(k = 1\): \(\binom{6}{1} x^5 \left( -\frac{2}{x} \right) = -12x^4\)

For \(k = 2\): \(\binom{6}{2} x^4 \left( -\frac{2}{x} \right)^2 = 60x^2\)

Thus, the first three terms are \(x^6 - 12x^4 + 60x^2\).

(ii) To find the coefficient of \(x^4\) in the expansion of \((1 + x^2) \left( x - \frac{2}{x} \right)^6\), we multiply the expansion from part (i) by \((1 + x^2)\):

\((1 + x^2)(x^6 - 12x^4 + 60x^2)\)

We need the terms that result in \(x^4\):

\(1 \times (-12x^4) = -12x^4\)

\(x^2 \times 60x^2 = 60x^4\)

Adding these gives \(-12 + 60 = 48\).

Thus, the coefficient of \(x^4\) is 48.