Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the distance between the two lenses is less than the focal length of any of them, the first image from the left lens will be to the right from the right lens.
Now we need to find the position of the first image from the left lens (lens 1). So we need to use the thin lens formula.
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
Thus,
$$s_1'=\left[ \dfrac{1}{f_1}-\dfrac{1}{s_1} \right]^{-1}$$
where $f_1=f_2=f$ and $s_1=\infty$ since the rays reach the lens parallel.
Therefore,
$$s_1'=f\tag 1$$
Now we need to find the second image where the author told us to use a negative object distance for the second lens, so $s_2=-(f-d)=d-f$ [see the figure below].
Using the thin lens formula again,
$$\dfrac{1}{f_2}=\dfrac{1}{s_2}+\dfrac{1}{s_2'}$$
Plugging the known and solving for $s_2'$,
$$\dfrac{1}{-f}=\dfrac{1}{d-f}+\dfrac{1}{s_2'}$$
$$s_2'=\left[ \dfrac{d-f-(-f)}{-f(d-f)}\right]^{-1}=\dfrac{-f(d-f)}{d }$$
$$s_2' =\dfrac{f^2-fd}{d }\tag 2$$
We need the effective focal length as the distance from the midpoint between the lenses to the final image. So
$$f_{\rm eff}=s_2'+\dfrac{d}{2}$$
Plugging from (2),
$$f_{\rm eff}=\dfrac{f^2-fd}{d }+\dfrac{d}{2}=\dfrac{2f^2-2fd+d^2}{2d}$$
Therefore,
$$\boxed{f_{\rm eff}= \dfrac{ f^2- fd+\frac{1}{2}d^2}{ d}}$$
$$\color{blue}{\bf [b]}$$
In this case, the zoom is given by
$$\textbf{Zoom}=\dfrac{f_{d_2}}{f_{d_1}}$$
where, according to the given data, $d_1=\frac{1}{2}f$ and $d_2=\frac{1}{4}f$
Using the boxed formula above,
$$\textbf{Zoom}=\dfrac{ \dfrac{ f^2- fd_2+\frac{1}{2}d_2^2}{ d_2}}{ \dfrac{ f^2- fd_1+\frac{1}{2}d_1^2}{ d_1}}$$
$$\textbf{Zoom}= \dfrac{ f^2- fd_2+\frac{1}{2}d_2^2}{ d_2} \dfrac{ d_1}{ f^2- fd_1+\frac{1}{2}d_1^2} $$
$$\textbf{Zoom}= \dfrac{ f^2- \frac{1}{4}f^2+\frac{1}{2}(\frac{1}{4}f)^2}{ \frac{1}{4} \color{red}{\bf\not} f} \dfrac{ \frac{1}{2} \color{red}{\bf\not} f}{ f^2- \frac{1}{2}f^2+\frac{1}{2}(\frac{1}{2}f)^2} $$
$$\textbf{Zoom}= \dfrac{2( f^2- \frac{1}{4}f^2+\frac{1}{32}f^2)} { f^2- \frac{1}{2}f^2+\frac{1}{8}f^2} $$
$$\textbf{Zoom}= \dfrac{2 \color{red}{\bf\not} f^2(1 - \frac{1}{4}+\frac{1}{32} )} { \color{red}{\bf\not} f^2(1- \frac{1}{2} +\frac{1}{8})} $$
$$\textbf{Zoom}=\color{red}{\bf2.5}\times $$
