Nice, but simple for problem 3. "$\Rightarrow$" $CEDF$ is cyclic , so $\angle{ACB}=\angle{BDF}=\angle{BDA}$, and now because $A,D,F$ are collinear we have that $\angle{BDA}=\angle{BDF}=90$, , therefore $E$ is the ortocenter of triangle $\triangle{FAB}$, thus $EF \perp AB$. "$\Leftarrow$" Denote $P \in AB \cap CD$, thus $EF$ is the polar of $P$ w.r.t to the circle, so $PO \perp EF$, where $O$ is the center of the circle. But $EF \perp AB$, and $P \in AB$, so $P,A,O,B$ are collinear, thus $\angle{ADB}=\angle{ACB}=\angle{BDF}=\angle{ACF}=90$, so $CEDF$ is cyclic.

$ \Rightarrow$ $ CEDF$ is cyclic, then $ \angle{ABD}=\angle{ACD}=\angle{EFD}=\alpha$, and $ \angle{BAC}=\angle{BDC}=\angle{EFC}=\beta$. Because $ ABCD$ is cyclic we also have that $ \angle{DBC}=\angle{CAD}=\gamma$. Since $ F$ is the intersection of $ AD$ and $ BC$, we have that $ \angle{BAF}+\angle{ABF}+\angle{AFB}=2\alpha+2\beta+2\gamma=180$. Therefore $ \alpha+\beta+\gamma=90$. Finally, we notice that $ \angle{ABF}+\angle{BFE}=\alpha+\gamma+\beta=90$. We conclude that $ EF\perp AB$. $ \Rightarrow EF\perp AB$. Since $ AB$ and $ CD$ are not parallel, we define $ G=AB\cap CD$. Since $ ABCD$ is cyclic, and $ G=AB\cap CD, E=AC\cap BD, F=AD\cap CB$, by Brocard's Theorem we can observe that $ O$ (center of the circle that pases through $ A,B,C,D$) is the orthocenter of $ \triangle{EFG}$. Considering that $ EF\perp AB$, or $ EF\perp GB$, we can observe that $ O$ has to be somewhere between $ A$ and $ B$ (because $ BG$ is one of the altitudes of $ \triangle{EFG}$, and naturally the orthocenter O must be on this line and on the circle containing $ A,B,C,D$, therefore $ O\in AB$). This leads to $ \angle{BCA}=\angle{BDA}=\angle{ECF}=\angle{EDF}=90$. Finally, we conclude that $ \angle{ECF}+\angle{EDF}=180$, and $ ECFD$ is cyclic. I have to agree with pohoatza. I doesn't seem like a problem 3.