The coefficient of \(x^2\) in \( (2-qx)^4 \) is

\( \binom42(2)^2(-q)^2=24q^2 \).

The coefficient of \(x^2\) in \( \left(1+\frac8q x\right)^6 \) is

\( \binom62\left(\frac8q\right)^2=\frac{960}{q^2} \).

Since the second expansion is subtracted,

\( 24q^2-\frac{960}{q^2}=324 \).

Multiply by \(q^2\):

\( 24q^4-324q^2-960=0 \).

Divide by \(12\):

\( 2q^4-27q^2-80=0 \).

Let \(u=q^2\). Then

\( 2u^2-27u-80=0 \).

\( (2u+5)(u-16)=0 \).

Since \(u=q^2\), \(u=16\).

Answer: \( q=4 \) or \( q=-4 \).