(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.