Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the speed is given by the distance over the time interval. So Yvette’s speed is then given y
$$v_{\rm y}=\dfrac{d}{ t}$$
where $d$ is the width of the room, where $d=n\left[\frac{1}{2}\lambda\right]$ since we know that each $\frac{1}{2}\lambda$ there is an antinodes where $n$ is the number of this antinodes.
Hence,
$$v_{\rm y}=\dfrac{n\left[\frac{1}{2}\lambda\right] }{ t}$$
Therefore, the number of sound maxima she hears per second is given by
$$\boxed{\dfrac{n}{t}=\dfrac{2v_{\rm y}}{\lambda}}$$
$$\color{blue}{\bf [b]}$$
According to Yvette's perspective, she is hearing a higher frequency $f_+$ from the source she is moving toward it, and at the same time, she is hearing a lower frequency $f_-$ from the opposite one she is moving away from it.
Hence, the beat frequency is given by
$$f_+-f_-=f\left[ 1+\dfrac{v_{\rm y}}{v} \right]-f\left[ 1-\dfrac{v_{\rm y}}{v} \right]$$
where $v$ is the speed of the wave.
$$f_+-f_-=f\left( 1+\dfrac{v_{\rm y}}{v} - 1+\dfrac{v_{\rm y}}{v} \right)$$
$$f_+-f_-=\dfrac{2v_{\rm y}}{v}f$$
where $v=v/\lambda$
$$\boxed{f_+-f_-=\dfrac{2v_{\rm y}}{\lambda}}$$
$$\color{blue}{\bf [c]}$$
As we can see from the two boxed formulas, we got the same results of beat numbers. Hence, the two answers are the same.
