Answer
See the detailed answer below.
Work Step by Step
Let's assume that at $t=0$, the three frames are overlapping at the origin point (0,0) of the $S$ frame. In other words, the three origins are at the same point at $t=0$ s.
$$\color{blue}{\bf [a]}$$
For frame $S'$;
As the author told us, we will use Lorentz transformation for space.
$$x'=\gamma(x-vt)$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.8^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{5}{3}$, and $v=0.8c$
Hence,
$$x'=\frac{5}{3}(x-0.8ct)$$
Plug the known;
$$x'=\frac{5}{3}(1200-[0.8\times 3\times 10^8][2\times 10^{-6}])$$
$$x'=\color{red}{\bf 1200}\;\rm m$$
Using Lorentz transformation for time.
$$t'=\gamma\left[t-\dfrac{vx}{c^2}\right] =\frac{5}{3}\left[t-\dfrac{0.8 cx}{c^2}\right] $$
$$t'= \frac{5}{3}\left[t-\dfrac{0.8 x}{c }\right] $$
Plug the known;
$$t'= \frac{5}{3}\left[(2\times 10^{-6})-\dfrac{0.8 (1200)}{(3\times 10^8) }\right] $$
$$t'=\color{red}{\bf -2.0}\;\rm \mu s$$
$$\color{blue}{\bf [b]}$$
For frame $S''$;
As the author told us, we will use Lorentz transformation for space.
$$x''=\gamma(x-vt)$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.8^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{5}{3}$, and $v=-0.8c$
Hence,
$$x''=\frac{5}{3}(x+0.8ct)$$
Plug the known;
$$x''=\frac{5}{3}(1200+[0.8\times 3\times 10^8][2\times 10^{-6}])$$
$$x''=\color{red}{\bf 2800}\;\rm m$$
Using Lorentz transformation for time.
$$t''=\gamma\left[t-\dfrac{vx}{c^2}\right] =\frac{5}{3}\left[t+\dfrac{0.8 cx}{c^2}\right] $$
$$t''= \frac{5}{3}\left[t+\dfrac{0.8 x}{c }\right] $$
Plug the known;
$$t''= \frac{5}{3}\left[(2\times 10^{-6})+\dfrac{0.8 (1200)}{(3\times 10^8) }\right] $$
$$t''=\color{red}{\bf 8.67}\;\rm \mu s$$
