Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to find the frequency difference between the third harmonic of the A and the second harmonic of the E, which is given by
$$f_{3A}-f_{2E}=3f_{1A}-2f_{1E}$$
Plugging the known;
$$f_{3A}-f_{2E}=3(440)-2(659)=\color{red}{\bf 2.0}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
We need to find the beat frequency between the first harmonics is given by
$$ f_{1E}-f_{1A}=659-440=\bf 219\;\rm Hz$$
The beat frequency between the second harmonics is given by
$$ f_{2E}-f_{2A}=2(659)-2(440)=\bf 438\;\rm Hz$$
And we know that the beat frequency between $f_{3A}$ and $f_{2E}$ is 2 Hz which means that he must listen for a beat frequency of 2 Hz.
$$\color{blue}{\bf [c]}$$
Since we got f a beat frequency of 4 Hz, then the second harmonic frequency of the E string is given by
$$f_{2E}=f_{3A}-4=3(440)-4=\bf 1316\;\rm Hz$$
Thus,
$$f_{1E}=\dfrac{f_{2E}}{2}=\dfrac{1316}{2}=\color{red}{\bf 658}\;\rm Hz$$
And since $f\propto \sqrt{T}$, the higher the tension the higher the frequency. Hence,
$$f_{2E}=f_{3A}+4=3(440)+4=\bf 1324\;\rm Hz$$
$$f_{1E}=\dfrac{f_{2E}}{2}=\dfrac{1324}{2}=\color{red}{\bf 662}\;\rm Hz$$
