Example 1
The subgroup $\{0,3\}$ of $\mathbb Z_6$ is a normal subgroup
I know it's a subgroup because the element $3\in\mathbb Z_6$ generates $\{0,3\}$ under addition
$\langle 3\rangle=(0,3)$ and the order is $2$ since we are back to the identity. But is it normal? I don't know how to prove/disprove it.
Example 2
The subgroup $\{e,h\}$ where $h$ is a reflection of $D_3$ is a normal subgroup.
Normally I would check if $\frac{|G|}{|H|}=2$ to show they are normal.
The usual test for normality of a subgroup $H \leq G$ is to check one of the following:
\begin{align*}gH = Hg \text{ for all }g \in G &\iff gHg^{-1} \subseteq H \text{ for all }g \in G \\ &\iff ghg^{-1} \in H \text{ for all }g \in G\text{ and }h \in H.\end{align*}
Of course, in an abelian group, the computations are all quite nice (for example, what is $ghg^{-1}$, only knowing that $g$ and $h$ belong to an abelian group?).
The result that $H \lhd G$ whenever $\lvert G : H\rvert = 2$ is a nice easy test for normality, but it too stems from the above criteria. You should keep the description normal subgroups are precisely those fixed by conjugation in mind.
If you're claiming that Example $2$ above is actually a normal subgroup, you should re-think that. Recall that $D_3 \cong S_3$, and think about what happens when you conjugate a $2$-cycle by anything.