Solution:

Here, $$d = 25 \,cm$$ $$f_0 = 8.0 \,mm, f_e = 2.5\, cm$$,

$$u_0 = -9.0\, mm = -0.9 \,cm$$

Now, $$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}} $$

$$ \therefore \frac{1}{u_{e}} = \frac{1}{v_{e}} -\frac{1}{f_{e}} = \frac{1}{-25}-\frac{1}{2.5}=\frac{-11}{25}$$

$$\left(\because v_{e} = -d = -25 \,cm\right) $$

$$ u_{e} = \frac{-25}{11}=2.27 \,cm $$

Again, $$\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}} $$

$$ \frac{1}{v_{0}} = \frac{1}{f_{0}} +\frac{1}{u_{0}} = \frac{1}{0.8} +\frac{1}{-0.9}$$

$$ \frac{0.9-0.8}{0.72} = \frac{0.1}{0.72} $$

$$ v_{0} = \frac{0.72}{0.1} = 7.2\, cm $$

Therefore, separation between two lenses

$$= u_{e} + v_{0} = 2.27 +7.2 = 9.47 \,cm$$

Magnifying power, $$m = \frac{v_{0}}{u_{0}}\left(1+\frac{d}{f_{e}}\right) $$

$$ = \frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right) = 88$$