Answer
(a) $d = f_2+f_1$
(b) $w_2 = \frac{f_2}{\vert f_1 \vert}\cdot w_1$
Work Step by Step
(a) In order for beams to exit the second lens as parallel beams, the focus point $f_1$ and the focus point $f_2$ must coincide at the same location a distance of $\vert f_1 \vert$ to the left of the first lens.
Therefore: $~~d = f_2- \vert f_1 \vert = f_2+f_1$
(b) We can use similar triangles to find $w_2$:
$\frac{w_2}{w_1} = \frac{f_2}{\vert f_1 \vert}$
$w_2 = \frac{f_2}{\vert f_1 \vert}\cdot w_1$
