(a) In the expansion of \((1-ax)^{\frac25}\), the coefficient of \(x^3\) is

\[ \frac{\frac25\left(\frac25-1\right)\left(\frac25-2\right)}{3!}(-a)^3. \]

So

\[ \frac{\frac25\left(-\frac35\right)\left(-\frac85\right)}{6}(-a)^3=1. \]

\[ \frac{8}{125}(-a)^3=1. \]

\[ -a=\frac52. \]

Therefore

\[ a=-\frac52. \]

(b) Since \(a=-\frac52\),

\[ (1-ax)^{\frac25}=\left(1+\frac52x\right)^{\frac25}. \]

The coefficient of \(x^3\) in this expansion is \(1\).

The coefficient of \(x^4\) in \(\left(1+\frac52x\right)^{\frac25}\) is

\[ \frac{\frac25\left(-\frac35\right)\left(-\frac85\right)\left(-\frac{13}{5}\right)}{4!}\left(\frac52\right)^4. \]

\[ =-\frac{13}{8}. \]

To get the coefficient of \(x^4\) in

\[ (2x+1)(1-ax)^{\frac25}, \]

we need:

\[ 1\times\text{coefficient of }x^4 \]

and

\[ 2x\times\text{coefficient of }x^3. \]

Hence the required coefficient is

\[ -\frac{13}{8}+2(1)=\frac38. \]

Answer: \(\frac38\).

(c) The binomial expansion is valid when

\[ \left|\frac52x\right|<1. \]

Therefore

\[ |x|<\frac25. \]

Answer: \(-\frac25<x<\frac25\).