Answer
a) $1.1\;\rm cm$
b) $160\times$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need the most powerful lens with the shortest focal length to be used as an objective lens.
Thus,
$$ f_{\rm obj}= \bf 1.0\;\rm cm$$
and
$$ f_{\rm eye}= \bf 2.0\;\rm cm$$
We are told that the distance between the two lenses is $L=16$ cm which means that the image formed by the first lens (the objective lens) must be at the focal point of the eye lens (viewing with a relaxed eye).
Hence, the first image from the objective lens must be at a distance of
$$s'_1=L-f_{\rm eye}=16-2=\bf 14\;\rm cm$$
Hence, the object position from the objective lens is given by the thin lens formula,
$$s_1=\left[\dfrac{1}{f_{\rm obj}}-\dfrac{1}{s'_1}\right]^{-1}$$
$$s_1=\left[\dfrac{1}{1}-\dfrac{1}{14}\right]^{-1}=\color{red}{\bf 1.1}\;\rm cm$$
$$\color{blue}{\bf [b]}$$
We know that the microscope magnification is given by
$$M=m_{\rm obj}M_{\rm eye}=\dfrac{-s_1'}{s_1}\dfrac{25\;\rm cm}{f_{\rm eye}}$$
We assumed that the lenses are simple magnifiers whereas $M=25\;{\rm cm}/f$, so
Plugging the known;
$$M= \dfrac{-14}{1.1}\cdot\dfrac{25 }{2}\approx\color{red}{ \bf-160}\times$$
