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

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

We need the term where the power of \(x\) is \(-3\):

\(x^{7-k} \left( -\frac{2}{x} \right)^k = x^{7-k} \cdot (-2)^k \cdot x^{-k} = (-2)^k x^{7-2k}\)

Set the exponent of \(x\) to \(-3\):

\(7 - 2k = -3\)

\(7 + 3 = 2k\)

\(10 = 2k\)

\(k = 5\)

Substitute \(k = 5\) into the binomial coefficient and expression:

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

\(= \binom{7}{5} x^{2-5} (-32)\)

\(= \binom{7}{5} x^{-3} (-32)\)

\(= 21 \cdot (-32) x^{-3}\)

\(= -672 x^{-3}\)

Thus, the coefficient of \(\frac{1}{x^3}\) is \(-672\).

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

\(\left( 3x + \frac{2}{3x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} (3x)^{7-k} \left( \frac{2}{3x^2} \right)^k\)

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

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

Set the power of \(x\) to \(-2\):

\(7 - 3k = -2\)

\(3k = 9\)

\(k = 3\)

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

\(\binom{7}{3} (3x)^{4} \left( \frac{2}{3x^2} \right)^3\)

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

\(= 35 \cdot 81 \cdot \frac{8}{27} \cdot x^{-2}\)

\(= 35 \cdot 3 \cdot 8 \cdot x^{-2}\)

\(= 840 \cdot x^{-2}\)

Thus, the coefficient of \(\frac{1}{x^2}\) is 840.

To find the coefficient of \(\frac{1}{x}\) in the expansion of \(\left( x - \frac{2}{x} \right)^5\), we use the binomial theorem:

\(\left( x - \frac{2}{x} \right)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left( -\frac{2}{x} \right)^k\)

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

\(x^{5-k} \cdot x^{-k} = x^{5-2k}\)

Set \(5-2k = -1\):

\(5 - 2k = -1\)

\(2k = 6\)

\(k = 3\)

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

\(\binom{5}{3} x^{5-3} \left( -\frac{2}{x} \right)^3\)

\(= \binom{5}{3} x^2 \left( -\frac{8}{x^3} \right)\)

\(= \binom{5}{3} \cdot (-8) \cdot x^{-1}\)

\(= 10 \cdot (-8) \cdot x^{-1}\)

\(= -80 \cdot x^{-1}\)

Thus, the coefficient of \(\frac{1}{x}\) is \(-80\).

(i) To find the coefficient of \(x^2\) in \((1 - 2x)^7\), use the binomial theorem: \(\binom{7}{2}(-2x)^2 = 21 \times 4x^2 = 84x^2\). The coefficient is 84.

For \(x^3\), \(\binom{7}{3}(-2x)^3 = 35 \times (-8x^3) = -280x^3\). The coefficient is -280.

(ii) The coefficient of \(x^3\) in \((2 + 5x)(1 - 2x)^7\) is found by expanding and combining terms: \(2 \times (-280) + 5 \times 84 = -560 + 420 = -140\).

(i) To find the coefficient of \(x\) in the expansion of \(\left(2x - \frac{1}{x}\right)^5\), we use the binomial theorem:

\(\left(2x - \frac{1}{x}\right)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} \left(-\frac{1}{x}\right)^k\)

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

\((2x)^{5-k} \left(-\frac{1}{x}\right)^k = 2^{5-k} (-1)^k x^{5-2k}\)

Set \(5-2k = 1\) to find \(k\):

\(5 - 2k = 1\)

\(2k = 4\)

\(k = 2\)

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

\(\binom{5}{2} (2x)^{3} \left(-\frac{1}{x}\right)^2 = 10 \cdot 8x^3 \cdot \frac{1}{x^2} = 80x\)

Thus, the coefficient of \(x\) is 80.

(ii) Now, find the coefficient of \(x\) in the expansion of \((1 + 3x^2) \left(2x - \frac{1}{x}\right)^5\).

We already know the coefficient of \(x\) in \(\left(2x - \frac{1}{x}\right)^5\) is 80.

We also need the coefficient of \(\frac{1}{x}\) in \(\left(2x - \frac{1}{x}\right)^5\):

Set \(5-2k = -1\):

\(5 - 2k = -1\)

\(2k = 6\)

\(k = 3\)

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

\(\binom{5}{3} (2x)^{2} \left(-\frac{1}{x}\right)^3 = 10 \cdot 4x^2 \cdot \left(-\frac{1}{x^3}\right) = -40 \cdot \frac{1}{x}\)

Now, combine these with \((1 + 3x^2)\):

Coefficient of \(x\) is:

\(1 \times 80 + 3 \times (-40) = 80 - 120 = -40\)

Thus, the coefficient of \(x\) is -40.