(a) The binomial expansion of \((a-x)^6\) is given by:

\((a-x)^6 = a^6 - 6a^5x + 15a^4x^2 - 20a^3x^3 + \ldots\)

Thus, the first four terms are \(a^6 - 6a^5x + 15a^4x^2 - 20a^3x^3\).

(b) Consider the expansion of \(\left(1 + \frac{2}{ax}\right)(a-x)^6\).

The term in \(x^2\) arises from:

\(\left(1 + \frac{2}{ax}\right)(15a^4x^2 - 40a^3x^3 + \ldots)\)

The coefficient of \(x^2\) is \(15a^4 - 40a^2\).

Given that this coefficient is \(-20\), we have:

\(15a^4 - 40a^2 = -20\)

Rearranging gives:

\(15a^4 - 40a^2 + 20 = 0\)

Factorizing gives:

\((5a^2 - 10)(3a^2 - 2) = 0\)

Solving these equations gives:

\(a^2 = 2\) or \(a^2 = \frac{2}{3}\)

Thus, \(a = \pm \sqrt{2}\) or \(a = \pm \frac{\sqrt{2}}{3}\).