Answer
a) $850\;\rm Hz$
b) $-\frac{1}{2}\pi$
Work Step by Step
$$\bf a)$$ First, we need to find the wavelength of the two sound waves.
$$\Delta \phi=\dfrac{2\pi \Delta x}{\lambda}+\Delta \phi_0$$
We can use the two constructive points at $x=0.50$ m and $x=0.90$ m.
Hence,
$$2\pi (m)=\dfrac{2\pi (0.5)}{\lambda}+\Delta \phi_0\tag 1$$
and
$$2\pi (m+1)=\dfrac{2\pi (0.9)}{\lambda}+\Delta \phi_0\tag 2$$
Subtracting (1) from (2);
$$2\pi (m+1)-2\pi (m)=\dfrac{2\pi (0.9)}{\lambda}+\Delta \phi_0 -\dfrac{2\pi (0.5)}{\lambda}+\Delta \phi_0$$
$$2\pi m+2\pi -2\pi m=\dfrac{2\pi (0.9)}{\lambda}+\Delta \phi_0 -\dfrac{2\pi (0.5)}{\lambda}-\Delta \phi_0$$
$$ 2\pi =\dfrac{2\pi (0.9)}{\lambda} -\dfrac{2\pi (0.5)}{\lambda} $$
$$ 1 =\dfrac{0.9}{\lambda} -\dfrac{0.5}{\lambda} =\dfrac{0.40}{\lambda}$$
Thus,
$$\lambda=\bf 0.40\;\rm m$$
So the frequency is then given by $v=\lambda f$, hence
$$f=\dfrac{v}{\lambda}=\dfrac{340}{0.40}=\color{red}{\bf 850}\;\rm Hz$$
$$\bf b)$$
The phase differnce between the two waves is given by (1);
$$2\pi (m)=\dfrac{2\pi (0.5)}{\lambda}+\Delta \phi_0 $$
where we use $m =1$,
$$2\pi =\dfrac{2\pi (0.5)}{0.4}+\Delta \phi_0 $$
Hence,
$$\Delta \phi_0=2\pi -\dfrac{2\pi (0.5)}{0.4}=2\pi\left[ 1-\dfrac{(0.5)}{0.4} \right]$$
$$\Delta \phi_0=\color{red}{\bf-\frac{1}{2}\pi} \;\rm rad$$
