Answer
a) $16.6\;\rm cm$
b) $5.4\times $
c) $77\%$
Work Step by Step
a) We know that the final image had to be at infinity for a relaxed eye. This means that the image of the first lens toward the object (the objective lens) must be at the focal length of the second lens (the eyepiece lens).
We know that the two lenses are having the same focal length of 12 cm.
Thus,
$$d_{i1}=d-f_1$$
where $d$ is the distance between the two lenses.
$$d_{i1}=55-12=\bf 43\;\rm cm$$
Now we can find the object position from
$$\dfrac{1}{f_1}=\dfrac{1}{d_{i1}}+\dfrac{1}{d_{o1}}$$
Solving for $d_{o1}$;
$$\dfrac{1}{f_1}-\dfrac{1}{d_{i1}}=\dfrac{1}{d_{o1}}$$
$$d_{o1}=\left[\dfrac{1}{f_1}-\dfrac{1}{d_{i1}} \right]^{-1}$$
Plugging the known;
$$d_{o1}=\left[\dfrac{1}{12}-\dfrac{1}{43} \right]^{-1}=\color{red}{\bf 16.6}\;\rm cm$$
-------------------
b) The magnification of the microscope is given by
$$M=M_eM_o=\dfrac{N(d-f_e)}{f_ed_o}$$
Plugging the known;
$$M =\dfrac{25(55-12)}{12\times 16.6}=\color{red}{\bf 5.4\times}$$
-------------------
c) We need first to apply the un-valid formula and then plug the result into the error percentage of
$${\rm Error\%}=\dfrac{M-M'}{M }\times 100\%$$
Thus,
$$M'=\dfrac{N l}{f_ef_o}$$
where $l$ here is $d$, as we see in the figure below.
$$M'=\dfrac{25\times 55 }{12\times 12}=\bf 9.55\times$$
Therefore,
$${\rm Error\%}=\bigg|\dfrac{5.4-9.55}{5.4}\bigg|\times 100\%=76.8\%\approx \color{red}{\bf 77\%}$$
