Problem number 10 seems to have a problem. The answer given assumes no phase shift from the reflection of the ray on the soap water interface. I think this is wrong. If Nwater>Nsoap>air both reflections result in a phase shift and so for a destructive interference 2 Nsoap x thickness = lambda/2. This gives lambda longer than the maximum for visible light. If one looks for the next destructive interfernce, it would require 2 Nsoap x thickness = 3 x lambda/2. This puts lambda at the edge or beow the visble light limit.
There may also be a problem for the oil drop problem,#13. I'll try to check it tonight.
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