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