Answer
(a) $f_2 \gt f_1$
(b) $f_2-f_1 = 1200~Hz$
Work Step by Step
(a) Since the speeder is approaching the police car, the speeder will detect a higher frequency $f_2$ than the frequency $f_1$ that the waves were emitted. That is, $f_2 \gt f_1$
(b) We can find $f_2-f_1$:
$\frac{f_2-f_1}{f_1} = \frac{v}{c}$
$f_2-f_1 = \frac{v}{c}~f_1$
$f_2-f_1 = \left(\frac{48.0~m/s}{3.0\times 10^8~m/s}\right)~(7.50\times 10^9~Hz)$
$f_2-f_1 = 1200~Hz$
