(i) The binomial expansion of \((a-x)^5\) is given by:

\(a^5 - 5a^4x + 10a^3x^2 - 10a^2x^3 + \ldots\)

Thus, the first four terms in ascending powers of \(x\) are \(a^5 - 5a^4x + 10a^3x^2 - 10a^2x^3\).

(ii) To find the coefficient of \(x^3\) in the expansion of \((1-ax)(a-x)^5\), consider the terms that contribute to \(x^3\):

The term from \((a-x)^5\) is \(-10a^2x^3\).

The term from \((1-ax)\) is \(-ax\), which when multiplied by the \(x^2\) term \(10a^3x^2\) from \((a-x)^5\), gives \(-10a^4x^3\).

Thus, the coefficient of \(x^3\) is:

\(-10a^4 - 10a^2 = -200\)

Simplifying, we get:

\(-10(a^4 + a^2) = -200\)

\(a^4 + a^2 = 20\)

Let \(y = a^2\), then:

\(y^2 + y - 20 = 0\)

Solving this quadratic equation using the quadratic formula:

\(y = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}\)

\(y = 4\) or \(y = -5\)

Since \(y = a^2\), \(y\) must be non-negative, so \(y = 4\).

Thus, \(a^2 = 4\), giving \(a = \pm 2\).