For Example 5, because of symmetry, if you can prove that $G',H',I',J'$ are concyclic, then by the same reason $F',H',I',J'$ are concyclic, and then of course all five points are on the same circle.
Angle chasing for $G',H',I',J'$ involves some other cyclic quadrilaterals. For example, try to prove that $A, B, F, G'$ are concyclic. Then by the same reason, so are $A, B, F, J'$. You can try to fill the details. We will discuss more in class,
For Example 6, instead of trying to prove $AEBF$ is a parallelogram, consider using properties of midpoints. We want to prove $O$ is the midpoint of $EF$. Draw a perpendicular line from $O$ to $CD$ with foot $G$, then $G$ is the midpoint of $CD$. Unfortunately points $G$ and $O$ are not in the same triangle, so we need to try to find another point to relate them... Again, discuss more in class.