Answer
See the detailed answer below.
Work Step by Step
Let's assume that at $t=0$, the two frames, $\rm S$ for Earth and $\rm S'$ for the train, are overlapping at the origin point (0,0) of the $\rm S$ frame which is for the observer on Earth. In other words, the two origins of the two frames are at the same point at $t=0$ s.
$$\color{blue}{\bf [a]}$$
Yes, the bell and siren are simultaneous events for a passenger on the train since the light starts from the train's center to its both ends and since the speed of light in the train's frame is independent of its speed.
$$\color{blue}{\bf [b]}$$
To answer this question, we need to find the space-time coordinates of the two events on frame $\rm S'$.
First, we need to find $t'$ which is the time on frame $\rm S'$ it takes the light to reach one end of the train.
$$t'=\dfrac{\frac{1}{2}L'}{c}$$
where $L'=30$ m which is the length of the train on frame $\rm S'$.
$$t'=\dfrac{ L'}{2c}=\dfrac{30}{2(3\times 10^8)}=\bf 0.05\;\rm \mu s$$
For the bell,
$$(x'_B,t'_B)=\rm (15\;m,0.05\;\mu s)\tag 1$$
For the siren,
$$(x'_s,t'_s)=\rm (-15\;m,0.05\;\mu s)\tag 2$$
Now we can use the Lorentz transformation formulas to find the times of these two events in frame $\rm S$ which is the bicyclist on the ground.
For the bell;
$$t_B=\gamma\left[t'_B+\dfrac{vx'_B}{c^2}\right] $$
where $v=0.5 c$, and $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.5^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{2\sqrt3}{3}$
Hence,
$$t_B= \frac{2\sqrt3}{3}\left[t'_B+\dfrac{0.5 cx'_B}{c^2}\right] $$
$$t_B= \frac{2\sqrt3}{3}\left[t'_B+\dfrac{0.5 x'_B}{c }\right] $$
Plug from (1),
$$t_B= \frac{2\sqrt3}{3}\left[(0.05\times 10^{-6})+\dfrac{0.5 (15)}{(3\times 10^8)}\right] $$
$$t_B=\bf 0.0866\;\rm\mu s$$
For the siren;
$$t_s=\gamma\left[t'_s+\dfrac{vx'_s}{c^2}\right] $$
$$t_s= \frac{2\sqrt3}{3}\left[t'_s+\dfrac{0.5 x'_s}{c }\right] $$
Plug from (1),
$$t_s= \frac{2\sqrt3}{3}\left[(0.05\times 10^{-6})+\dfrac{0.5 (-15)}{(3\times 10^8)}\right] $$
$$t_s=\bf 0.0289\;\rm\mu s$$
Now it is obvious that the siren event occurs before the bell event by a time of
$$\Delta t= t_B-t_s=0.0866-0.0289=\color{red}{\bf 0.0577}\;\rm \mu s$$
