Step 1: Find the object distance for the eyepiece.
Focal length of the objective lens, $f_{\circ }$​= 8 mm = 0.8 cm
The focal length of the eyepiece, $f_e$​= 2.5 cm
Object distance for the objective lens, $u_{\circ }$​= -9.0 mm = -0.9 cm
Least distance of distant vision, d=25 cm
Image distance for the eyepiece, $v_e$​= -d=-25 cm
Using the lens formula,

$$$\begin{array}{l}\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\\ \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\\ =\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25}\\ \therefore u_e=-\frac{25}{11}=-2.27 cm\end{array}$

Step 2: Find the image distance for the objective lens.
For the objective lens, using the lens formula :

$\begin{array}{l}\frac{1}{v_{\circ }}-\frac{1}{u_{\circ }}=\frac{1}{f_{\circ }}\\ \frac{1}{v_{\circ }}=\frac{1}{f_{\circ }}+\frac{1}{u_{\circ }}\\ =\frac{1}{0.8}-\frac{1}{0.9}=\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}\\ \therefore v_{\circ }=7.2 cm\end{array}$

Step 3: Find the distance between the lens.
The distance between the objective lens and the eyepiece $=\left|u_e\right|+v_{\circ }$=2.27 + 7.2= 9.47 cm

Step 4: Find the magnifying power of the microscope.
The magnifying power of the microscope,

$m=\frac{v_{\circ }}{\left|u_{\circ }\right|}\left(1+\frac{d}{f_e}\right)$
$=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=8\left(1+10\right)=88$
Hence, the magnifying power of the microscope is 88.