Answer
a) $49.5\;\rm mm$
b) $2.2\;\rm cm$
Work Step by Step
As the author described the camera's normal lens, the image is extended on the full of the film or the screen of the camera when the object is very far away from the camera in which way it makes a horizontal angle of 40$^\circ$, as we can see below.
This means that the object distance is much greater than the focal length of the lens.
$$d_o\gt\gt f$$
Thus, the image distance is given by
$$\dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o}$$
$$\dfrac{1}{f}- \dfrac{1}{d_o}=\dfrac{1}{d_i}$$
Solving for $d_i$;
$$d_i=\dfrac{fd_o}{d_o-f }$$
and since $(d_o\gt\gt f)$, so $(d_o-f\approx d_o)$
$$d_i=\dfrac{fd_o}{d_o }=f$$
Therefore, the image is just at the focal length of the camera.
From the geometry of the figure below, we can see that
$$\tan20^\circ=\dfrac{\frac{1}{2}L}{f}$$
Thus, the focal length is given by
$$\boxed{f=\dfrac{\frac{1}{2}L}{\tan20^\circ}}$$
a) Now we need to find the focal length of the 35-mm camera.
Using the boxed formula above;
$$f=\dfrac{\frac{1}{2}L}{\tan20^\circ}=\dfrac{\frac{1}{2}\times 36}{\tan20^\circ}=\color{red}{\bf 49.5}\;\rm mm$$
b) And we also need to find the focal length of the digital camera.
Using the same boxed formula above;
$$f=\dfrac{\frac{1}{2}L}{\tan20^\circ}=\dfrac{\frac{1}{2}\times 1.5}{\tan20^\circ}=\color{red}{\bf 2.2}\;\rm cm$$
