(i) To find the coefficient of \(x^2\) in \(\left( 2x - \frac{1}{2x} \right)^6\), we use the binomial expansion:

The general term is \(\binom{6}{r} (2x)^{6-r} \left(-\frac{1}{2x}\right)^r\).

We need \((6-r) - r = 2\), which gives \(6 - 2r = 2\), so \(r = 2\).

The term is \(\binom{6}{2} (2x)^4 \left(-\frac{1}{2x}\right)^2\).

Calculating, \(\binom{6}{2} = 15\), \((2x)^4 = 16x^4\), and \(\left(-\frac{1}{2x}\right)^2 = \frac{1}{4x^2}\).

The coefficient of \(x^2\) is \(15 \times 16 \times \frac{1}{4} = 60\).

(ii) For \((1 + x^2) \left( 2x - \frac{1}{2x} \right)^6\), we consider two terms:

1. The constant term from \(\left( 2x - \frac{1}{2x} \right)^6\) is \(20 \times x^3 \times \left(-\frac{1}{2x}\right)^3 = 20 \times 8x^3 \times \left(-\frac{1}{8x^3}\right) = 20\).

2. The \(x^2\) term from \((1 + x^2)\) is \(60\) from part (i).

The coefficient of \(x^2\) is \(60 - 20 = 40\).

To find the coefficient of \(x^3\) in the expansion of \(\left( 2 - \frac{1}{2}x \right)^7\), we use the binomial theorem:

\(\binom{n}{k} a^{n-k} b^k\)

where \(n = 7\), \(a = 2\), \(b = -\frac{1}{2}x\), and we need the term where the power of \(x\) is 3.

Thus, \(k = 3\) and the term is:

\(\binom{7}{3} (2)^{7-3} \left(-\frac{1}{2}x\right)^3\)

Calculate each part:

\(\binom{7}{3} = 35\)

\((2)^4 = 16\)

\(\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\)

Combine these to find the coefficient:

\(35 \times 16 \times -\frac{1}{8} = -70\)

Therefore, the coefficient of \(x^3\) is \(-70\).

To find the coefficient of \(x^6\) in the expansion of \(\left( 2x^3 - \frac{1}{x^2} \right)^7\), we use the binomial theorem:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

Here, \(a = 2x^3\) and \(b = -\frac{1}{x^2}\), and \(n = 7\).

We need the term where the power of \(x\) is 6. The general term is:

\(T_k = \binom{7}{k} (2x^3)^{7-k} \left(-\frac{1}{x^2}\right)^k\)

Simplifying, we get:

\(T_k = \binom{7}{k} \cdot 2^{7-k} \cdot x^{3(7-k)} \cdot (-1)^k \cdot x^{-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-3k-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-5k}\)

We want \(21-5k = 6\), so:

\(21 - 5k = 6\)

\(5k = 15\)

\(k = 3\)

Substitute \(k = 3\) into the term:

\(T_3 = \binom{7}{3} \cdot 2^{4} \cdot (-1)^3 \cdot x^{6}\)

\(= 35 \cdot 16 \cdot (-1) \cdot x^{6}\)

\(= -560 \cdot x^{6}\)

Thus, the coefficient of \(x^6\) is \(-560\).

To find the coefficient of \(x\) in the expansion of \(\left( x + \frac{2}{x^2} \right)^7\), we use the binomial theorem:

\(\left( x + \frac{2}{x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \left( \frac{2}{x^2} \right)^k\)

We need the term where the power of \(x\) is 1:

\(7-k - 2k = 1\)

Solving for \(k\):

\(7 - 3k = 1\)

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) into the binomial term:

\(\binom{7}{2} x^{7-2} \left( \frac{2}{x^2} \right)^2 = \binom{7}{2} x^5 \left( \frac{4}{x^4} \right)\)

\(= \binom{7}{2} \cdot 4 \cdot x = 21 \cdot 4 \cdot x = 84x\)

Thus, the coefficient of \(x\) is 84.

To find the coefficient of \(x^2\) in the expansion of \(\left( \frac{x}{2} + \frac{2}{x} \right)^6\), we use the binomial theorem.

The general term in the expansion is given by:

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

We need the term where the power of \(x\) is 2:

\((6-k) - k = 2\)

\(6 - 2k = 2\)

\(2k = 4\)

\(k = 2\)

Substitute \(k = 2\) into the general term:

\(T_2 = \binom{6}{2} \left( \frac{x}{2} \right)^4 \left( \frac{2}{x} \right)^2\)

\(= 15 \left( \frac{x^4}{16} \right) \left( \frac{4}{x^2} \right)\)

\(= 15 \times \frac{x^4}{16} \times \frac{4}{x^2}\)

\(= 15 \times \frac{x^2}{4}\)

\(= \frac{15}{4} x^2\)

Thus, the coefficient of \(x^2\) is \(\frac{15}{4}\) or 3.75.