Answer
a) $\approx 82\;\rm nm$
b) $\approx 130\;\rm nm$
Work Step by Step
As we see in the figure below, the two reflected rays experience a $\pi$ phase change due to reflection from a medium with a greater index of refraction.
For destructive interference from the film when the two rays are reflected in phase, the path length difference is given by
$$2t=\left(m+\frac{1}{2}\right)\lambda_{n }$$
$$2t=\dfrac{\left(m+\frac{1}{2}\right)\lambda }{n_{film}}$$
And hence the thickness of the film is given by
$$ t=\dfrac{\left(m+\frac{1}{2}\right)\lambda }{2n_{film}}$$
And for the minimum thickness, $m=0$
$$ t=\dfrac{\left(0+\frac{1}{2}\right)\lambda }{2n_{film}}$$
$$ t=\dfrac{ \lambda }{4n_{film}}$$
a)
For the blue light
$$ t=\dfrac{ \lambda }{4n_{film}}=\dfrac{450}{4\cdot 1.38}=\color{red}{\bf 81.52}\;\rm nm$$
b)
For the red light
$$ t=\dfrac{ \lambda }{4n_{film}}=\dfrac{720}{4\cdot 1.38}=\color{red}{\bf 130.4}\;\rm nm$$
