Answer
a) $4\;\rm m/s$
b) $8\;\rm m$
Work Step by Step
Let's assume that at $t=0$, frame $S'$ overlaps with frame $S'$.
$$\color{blue}{\bf [a]}$$
Now we need to use the Galilean transformations of position to find $x_1'$ and $x_2'$;
$$x=x'+vt$$
So, the first explosion at $t_1$ occurs at the position of
$$x=x_1' +vt_1$$
Plug the known;
$$x_1 =4+v\tag 1$$
And the second explosion at $t_2$ occurs at the position of
$$x_2=x_2'-vt_2$$
Plug the known;
$$x_2=-4+3v\tag 2$$
where $x_1=x_2$ since the two events occur in the same place but at different times.
Solving (1) and (2);
$$4+v=-4+3v$$
$$v=\color{red}{\bf 4}\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
Plug the velocity into (1) to find $x$,
$$x=\color{red}{\bf8}\;\rm m$$
