Answer: \(a^2b=-12\), \(a=2\), \(b=-3\).

Use the binomial theorem carefully, keeping track of the power of each factor in the general term.

The general term in the expansion is

\(T_{r+1}=\binom9r(ax)^{9-r}\left(\frac{b}{x^2}\right)^r.\)

So

\(T_{r+1}=\binom9r a^{9-r}b^r x^{9-r-2r} =\binom9r a^{9-r}b^r x^{9-3r}.\)

For the term independent of \(x\),

\(9-3r=0,\)

so \(r=3\). Its coefficient is

\(\binom93a^6b^3=84a^6b^3.\)

This is given as \(-145\,152\), so

\(84a^6b^3=-145\,152.\)

Hence

\(a^6b^3=-1728.\)

Therefore

\((a^2b)^3=-1728,\)

and so

\(a^2b=-12.\)

For the coefficient of \(x^6\),

\(9-3r=6,\)

so \(r=1\). The coefficient is

\(\binom91a^8b=9a^8b.\)

Since this coefficient is \(-6912\),

\(9a^8b=-6912.\)

Using \(a^2b=-12\), we have \(b=-\frac{12}{a^2}\). Substitute this into \(9a^8b=-6912\):

\(9a^8\left(-\frac{12}{a^2}\right)=-6912.\)

Thus

\(-108a^6=-6912,\)

so

\(a^6=64.\)

Since \(a\gt 0\),

\(a=2.\)

Finally,

\(b=-\frac{12}{a^2}=-\frac{12}{4}=-3.\)