Answer
See the detailed answer below.
Work Step by Step
a)
In this situation, we need to apply Doppler's effect twice.
First, when the source is getting closer, and second when it reflects the sound.
We know when the object is getting closer
$$f_+=f_0\left[1+\dfrac{v_0}{v}\right]\tag 1$$
And when the sound is reflected
$$f_+'=\dfrac{f_+}{1-\dfrac{v_0}{v}} $$
Plugging $f_+$ from (1);
$$f_+'=\dfrac{f_0\left[1+\dfrac{v_0}{v}\right]}{1-\dfrac{v_0}{v}} =\left[\dfrac{\dfrac{v+v_0}{v}}{\dfrac{v-v_0}{v}}\right]f_0$$
$$f_+'=\dfrac{v+v_0}{v-v_0}f_0$$
Thus,
$$\Delta f=f_+'-f_0=\dfrac{v+v_0}{v-v_0}f_0-f_0$$
$$\Delta f =\left[\dfrac{v+v_0}{v-v_0} -1\right]f_0=\left[\dfrac{v+v_0-v+v_0}{v-v_0} \right]f_0$$
$$\Delta f = \left[\dfrac{2v_0}{v-v_0} \right]f_0$$
and since $v_0\lt\lt v$, so
$$\boxed{\Delta f = \dfrac{2f_0v_0}{v } }$$
---
b) According to Bernoulli’s equation for fluid dynamics
$$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
The author told us that the blood inside the heat is at rest, so
$v_1=0$ and also he told us that the patient is lying down, so $y_1=y_2$.
$$P_1 =P_2+\frac{1}{2}\rho v_2^2 $$
Noting that $v_2$ is actually $v_0$ in the boxed formula above.
Hence,
$$P_2-P_1 =-\frac{1}{2}\rho v_0^2 =-\frac{1}{2}\rho \left[ \dfrac{\Delta f v}{2f_0}\right]^2 $$
$$\Delta P =- \frac{1}{2}\rho \left[ \dfrac{\Delta f v}{2f_0}\right]^2 $$
Plugging the given;
$$\Delta P = \frac{1}{2}(1060)\left[ \dfrac{(6000) (1480)}{2 (2.5\times 10^6)}\right]^2 =-\bf1672\;\rm Pa$$
Noting that we assumed that the speed of sound in blood is equal to the speed of sound in water.
$$\Delta P =\color{red}{\bf 12.54}\;\rm mmHg$$