Answer
See explanation.
Work Step by Step
a. The sound will be maximally reinforced when the path difference is an integer multiple of wavelengths.
Constructive interference occurs for $2.04 \mu m=2040 nm =0, \lambda, 2\lambda, 3\lambda, 4\lambda, …$
The values of $\lambda$ in the visible range are $\frac{2040 nm}{3}=680 nm$, $\frac{2040 nm}{4}=510 nm$, and $\frac{2040 nm}{5}=408 nm$.
b. The path-length difference is the same as before. The wavelengths are the same as before.
c. The sound will be canceled when the path difference is an odd multiple of half wavelengths.
Destructive interference occurs for $2040nm=\lambda/2, 3\lambda/2, 5\lambda/2, 7\lambda/2, …$
The values of $\lambda$ in the visible range are $\frac{2 \times 2040 nm}{7}=583 nm$, and $\frac{2 \times 2040 nm}{9}=453 nm$.
The wavelengths for destructive interference fall between the ones for constructive interference.
