The idea is that yes you multiply by 1, but remember that t is approaching zero which means that the limit has to be equal to 1 for these terms and in case it is.
If these new limits were not equal to 1 then you couldn't do the substitution.
Yes that is exactly the idea. Though I would simply do this...
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{3x}}{\sin{6x}} &= \lim_{x \to 0}\frac{\sin{3x}}{2\sin{3x}\cos{3x}} \\ &= \lim_{x \to 0}\frac{1}{2\cos{3x}} \\ &= \frac{1}{2 \cos {0}} \\ &= \frac{1}{2 \cdot 1} \\ &= \frac{1}{2} \end{align*}$