Answer
$145\;\rm m/s,\;15\;m/s$
Work Step by Step
Let's assume that the sprinter is moving from left to right. So we have here two references of frames, the sprinter frame $S'$ and the air which is the medium at which the sound moves through it (where air here is the stationary frame $S$).
Let's assume that $v_1,v_1'$ are the speeds of sound from in front crowding, while $v_2,v_2'$ are the speeds of sound from behind crowding. See the figure below.
Now we need to use the Galilean transformations of velocity.
For the in-front sound, from the figure below,
$$v_1'=v_1-v=-v_{\rm sound}-v$$
Plug the known;
$$ -360=-v_{\rm sound}-v\tag 1$$
For the back sound, from the figure below,
$$v_2'=v_2-v=v_{\rm sound}-v$$
Plug the known;
$$ 330=-v_{\rm sound}-v\tag 2$$
(1) plus (2);
$$-30=-2v$$
Thus, the velocity of the sprinter is
$$v=\color{red}{\bf 15}\;\rm m/s$$
Plug into (1) to find the speed of the sound;
$$ -360=-v_{\rm sound}-15$$
$$ v_{\rm sound}=360-15$$
$$ v_{\rm sound}=\color{red}{\bf 345}\;\rm m/s$$
