Answer
See the detailed answer below.
Work Step by Step
We have a film of oil on a glass piece, where we know that the oil's index of refraction is less than that of the glass and greater than that of the air.
So we have two 180$^\circ$ phase changes, meaning that the two reflected rays will interact in phase, as we see in the figure below.
Hence, for constructive interference,
$$2t=m\dfrac{\lambda}{n_{film}}\tag 1$$
where $t$ is the thickness of the film.
And for destructive interference,
$$2t=\left(m+\frac{1}{2}\right)\dfrac{\lambda}{n_{film}}\tag 2$$
$$\color{blue}{\bf a)}$$
Solving (1) for $\lambda$, to find the visible wavelengths of light that reflected and interfere constructively.
$$\lambda=\dfrac{2tn_{film}}{m}$$
Plugging the known, and recalling that the range of the visible light is from 380 nm to 700 nm.
$$\lambda=\dfrac{2(500)(1.42)}{m}=\dfrac{1420}{m}$$
At $m=1$,
$$\lambda=\bf 1420\;\rm nm$$
which is not visible light.
At $m=2$,
$$\lambda=\bf 710\;\rm nm$$
which is not visible light.
At $m=3$,
$$\lambda=\color{red}{\bf 473}\;\rm nm$$
which is visible light.
At $m=4$,
$$\lambda=\bf 355\;\rm nm$$
which is not visible light.
So, it is obvious that the only visible light reflected constructively is 473 nm.
$$\color{blue}{\bf b)}$$
Solving (2) for $\lambda$, to find the visible wavelengths of light that reflected and interfere destructively.
$$\lambda=\dfrac{2tn_{film}}{m+\frac{1}{2}} $$
Plugging the known,
$$\lambda=\dfrac{1420}{m+\frac{1}{2}} $$
At $m=1$,
$$\lambda=\bf 947\;\rm nm$$
which is not visible light.
At $m=2$,
$$\lambda=\color{red}{\bf 568}\;\rm nm$$
which is visible light.
At $m=3$,
$$\lambda=\color{red}{\bf 406}\;\rm nm$$
which is visible light.
At $m=4$,
$$\lambda=\bf 316\;\rm nm$$
which is not visible light.
So, it is obvious that the only visible light reflected destructively is $568$ nm, $406$ nm.
$$\color{blue}{\bf c)}$$
The color of the reflected light is 473 nm light which is blue.
And the transmitted lights are 406 nm which is violet and 568 nm which is yellow-green light.
