Answer
See the detailed answer below.
Work Step by Step
First, we need to sketch the problem, as seen below.
$$\color{blue}{\bf [a]}$$
We know that the phase difference is given by
$$\Delta \phi=\dfrac{2\pi \Delta r}{\lambda}+\Delta \phi_0$$
where $\Delta \phi_0=0$ since the courses are identical, and $\lambda=c/f$ where $c$ is the speed of the electromagnetic waves.
$$\Delta \phi=\dfrac{2\pi f \Delta r}{c}\tag 1$$
Now we need to find $\Delta r$ which is given by
$$\Delta r=r_2-r_1$$
$$\Delta r=\sqrt{(600-[-50])^2+(800-0)^2}-\sqrt{(600-50)^2+(800-0)^2}=\bf 60\;\rm m$$
Plugging all the known into (1);
$$\Delta \phi=\dfrac{2\pi (3\times 10^6)(60)}{(3\times 10^8)} $$
$$\Delta \phi=\frac{6}{5}\pi=\color{red}{\bf 1.2\pi}\;\rm rad$$
$$\color{blue}{\bf [b]}$$
To find the type of interference, we need to find the wavelength and compare it with the path difference between these two waves.
If the path difference was an integer of $\lambda$, it is a perfect constructive interference. And if it is an integer of $\frac{1}{2}\lambda$, it is a perfect constructive interference. If neither that nor this, it is somewhere in between.
$$\lambda=\dfrac{c}{f}=\dfrac{3\times 10^8}{3\times 10^6}=\bf 100\;\rm m$$
So,
$$\dfrac{\Delta r}{\lambda}\lambda=\dfrac{60}{100}\lambda=0.6\lambda$$
Therefore, it is $\color{red}{\bf {\text{ somewhere in between}}}$.
$$\color{blue}{\bf [c]}$$
As we move north, the path length difference increases and hence the phase difference increases until we reach a point at which $\Delta \phi =2\pi$ where the interference there is a perfect constructive one.
Therefore, the signal strength$\color{red}{\bf {\text{ increases}}}$.
