Finding the maximum phase and the maximum frequency deviation of an angle-modulated signal
I think this is treating the $10^8\pi\cdot t$ as the carrier frequency and the $5 sin(2\pi\cdot 10^3 t)$ as the modulation.  So the maximum phase difference of this signal to the carrier without modulation would be $\pm5$ radians.
To compute the change in frequency you need to find the compute $\frac{d}{dt}$ of the argument of the cos (divided by $2\pi$) and subtract the carrier frequency.
$$\frac{d}{dt} 10^8\pi\cdot t + 5 sin(2\pi\cdot 10^3 t) = 10^8\pi + 2\cdot 5\cdot 10^3\pi cos(2\pi\cdot 10^3 t)  $$
Divide by $2\pi$ to get Hz instead of radians.
$$Frequency = 5\cdot 10^7  + 5\cdot 10^3 cos(2\pi \cdot 10^3t)$$
Subtract off the carrier of 50MHz and you have a maximum frequency excursion of $\pm 5$ k Hz