Answer
a) $5\;\rm beats/s$
b) $4.545\;\rm mm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to find the new frequency when the air speed changes.
$$f'=\dfrac{v'}{\lambda}$$
And since she plays the same note A as the tuning fork, the wavelength does not change.
Hence,
$$\lambda=\dfrac{v}{f}$$
Plugging into the first formula above;
$$f'=\dfrac{v'}{v}f\tag 1$$
Now the number of beats is given by
$$f_{beats}=f'-f\tag{Plug from (1)}$$
$$f_{beats}=\dfrac{v'}{v}f-f=\left[\dfrac{v'}{v}-1\right]f$$
Plugging the known;
$$f_{beats}= \left[\dfrac{346}{342}-1\right]440=5.14\;\rm Hz$$
$$f_{beats}=\color{red}{\bf 5}\;\rm beats/s$$
$$\color{blue}{\bf [b]}$$
We know that the initial length of the flute is given by
$$L=\dfrac{\lambda}{2}=\dfrac{v}{2f}\tag 2$$
And hence, the final length is then
$$L'= \dfrac{v'}{2f}\tag 3$$
Thus the extension is given by
$$\Delta L= L'-L$$
Plugging from (2) and (3);
$$\Delta L= \dfrac{v'}{2f}- \dfrac{v}{2f}= \dfrac{v'-v}{2f}$$
Plugging the known;
$$\Delta L = \dfrac{346-342}{2(440)}=\bf 4.54545\times 10^{-3}\;\rm m$$
$$\Delta L=\color{red}{\bf 4.545}\;\rm mm$$
