Answer: \(n=6\), \(a=5\), \(b=60\).

The first term of \((2-ax)^n\) is \(2^n\). Comparing with the given expansion,

\(2^n=64.\)

So

\(n=6.\)

Now use the binomial expansion of \((2-ax)^6\).

The coefficient of \(x\) is

\(\binom61 2^5(-a)=6\cdot 32(-a)=-192a.\)

This must equal \(-16b\), so

\(-192a=-16b \quad\Rightarrow\quad b=12a.\)

The coefficient of \(x^2\) is

\(\binom62 2^4 a^2=15\cdot 16a^2=240a^2.\)

This must equal \(100b\), so

\(240a^2=100b.\)

Substitute \(b=12a\):

\(240a^2=1200a.\)

Thus

\(a(a-5)=0.\)

Since the \(x\)-term is present, \(a\neq 0\). Therefore

\(a=5.\)

Then

\(b=12a=60.\)

Hence

\(\boxed{n=6,\quad a=5,\quad b=60}.\)