$S^2$ is rotation by ${180}^{\circ }$ about the origin. $S^3=S^{-1}$ is rotation by ${270}^{\circ }$ clockwise about the origin. $S^4=I$. So $S^{2006}=S^2$.

$T^2$ is translation two units to the right; $T^3$ is translation three units to the right, and so on. $T^{2006}$ is translation 2006 units to the right.

$RS$ is not the same as $SR$. (In fact, to get from one to the other you could rotate by ${180}^{\circ }$ about the origin - but this isn't a surprise, as to get from the image of $RS$ to the image of $SR$ we'd do $S^{-1}{R}^{-1}$ (the inverse of $RS$) followed by $SR$, i.e., $(S^{-1}{R}^{-1})(SR)=S^{-1}(R^{-1}SR)=S^{-1}{S}^{-1}=S^{-2}=S^2$. However, the relationship $R^{-1}SR=S^{-1}$ isn't introduced until the 3 star problem!).
$R{T}^2$ is not the same as $T^{2}R$.
$(RT)S$ is not the same as $S(RT)$.
However, changing the order does sometimes produce the same transformation: $R{S}^2=S^{2}R$, for example. (Of course, $I$ commutes with everything!)

The inverse of $RS$ is $(RS{)}^{-1}=S^{-1}{R}^{-1}$.

$(ST{)}^{-1}=T^{-1}{S}^{-1}$.

$(TR{)}^{-1}=R^{-1}{T}^{-1}$.

$(R{S}^2{)}^{-1}=S^{-2}{R}^{-1}$.