(a) To find the coefficient of \(x^2\) in \(\left(x - \frac{2}{x}\right)^6\), we use the binomial expansion. The general term is given by:

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

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

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

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

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

Thus, the coefficient of \(x^2\) is 60.

(b) To find the coefficient of \(x^2\) in \((2 + 3x^2)\left(x - \frac{2}{x}\right)^6\), we first find the constant term in \(\left(x - \frac{2}{x}\right)^6\):

The constant term is when the power of \(x\) is 0:

\(6-k-k = 0 \Rightarrow k = 3\).

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

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

Now, consider the expansion \((2 + 3x^2)(-160)\):

The coefficient of \(x^2\) is:

\(3x^2(-160) = -480\).

Adding the contribution from \(2\) times the \(x^2\) term from part (a):

\(2 \times 60 = 120\).

Thus, the total coefficient of \(x^2\) is:

\(120 - 480 = -360\).