The given expression is $\left(a x-\frac{1}{b x^2}\right)^{13}$

So,

$$ \begin{aligned} T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\ & ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\ & ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r} \end{aligned} $$

For coefficient of $x^7$ in $\left(a x-\frac{1}{b x^2}\right)^{13}$

$$ \begin{aligned} & \quad 13-3 r=7 \\\\ & \Rightarrow 3 r=6 \Rightarrow r=2 \\\\ & \therefore \text { Coefficient of } x^7={ }^{13} C_2 \cdot(a)^{11} \cdot \frac{1}{b^2} \end{aligned} $$

Again, the another expression is $\left(a x+\frac{1}{b x^2}\right)^{13}$

So, $T_{n+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r={ }^{13} C_r(a)^{13-r}\left(\frac{1}{b}\right)^r x^{13-3 r}$

For coefficient $x^{-5}$ in $\left(a x+\frac{1}{b x^2}\right)^{13}$

$$ \begin{aligned} &13-3 r =-5 \\\\ &\Rightarrow r =6 \end{aligned} $$

So, coefficient of $x^{-5}={ }^{13} C_6(a)^7 \frac{1}{b^6}$

Now, according to the question,

${ }^{13} C_2(a)^{11} \frac{1}{b^2}={ }^{13} C_6(a)^7 \frac{1}{b^6}$

$$ \begin{aligned} & \Rightarrow a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2} \\\\ & \therefore a^4 b^4=22 \end{aligned} $$