$1)$: I used coordinate geometry, and it actually made things really easy. Set $D$ to be the origin: $(0,0)$. Then, notice that $\Delta ABC, \Delta ABD,$ and $\Delta ADC$ are $30-60-90$ triangles. Also, set the length of $AE=a$, so that $DE=3a$ and $AD=4a$. Now, we can deduce some coordinates: $B=(4a\sqrt{3},0), C=(-\frac{4a\sqrt{3}}{3},0), A=(0,4a)$, and $E=(0,3a)$. To prove that $AF \perp FC$, we can show that, if $m_{XY}$ denotes the slope of the line $XY$, $m_{AF}=-\frac{1}{m_{CF}}$. To do so, we first find the coordinates of $F$ by finding the equations of lines $DF$ and $BE$. $DF: y=\frac{4}{\sqrt{3}}x; BE: y=-\frac{\sqrt{3}}{4}+3a$. Solving these equations gives us $F=(\frac{12a\sqrt{3}}{19},\frac{48a}{19})$. Now, we can easily compute the slopes of lines $AF$ and $CF$. $m_{AF} = -\frac{7}{3\sqrt{3}}$, and $m_{CF} = \frac{3\sqrt{3}}{7}$. Hence, since $m_{AF}=-\frac{1}{m_{CF}}$, we have shown that $AF \perp FC$. $2)$: We can use our coordinates we deduced from the previous section. Focus on $\Delta AFB$. We can find the side lengths $AF, FB,$, and $BA$, and then use Law of Cosines to find $m\angle AFB$. By the distance formula, we have $FB=\frac{16a\sqrt{57}}{19}, AF = \frac{8a\sqrt{19}}{19}$, and $AB=8a$. We can use now use Law of Cosines: $\cos \angle AFB = \frac{AF^{2}+FB^{2}-AB^{2}}{2 \times AF \times FB}=\frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$. We can then deduce that $m\angle AFB=arccos(-\frac{\sqrt{3}}{2})=\boxed{150º}$.

We show that $\triangle FEA \sim \triangle FDC$ Since $\angle FED = \angle FDB$, Therefore $\angle FEA = \angle FDC$. Also, $$\frac{FE}{EA}=\frac{1}{3}\frac{FE}{DE}=\frac{1}{3}\frac{FD}{BD}=\frac{FD}{DC}$$Hence, $\triangle FEA \sim \triangle FDC$. Next, $\angle AFE = \angle DFC \implies \angle AFC = \angle EFD = 90^{\circ}$ Also, we have $AFDC$ cyclic. Hence, $\angle BFA = 180^{\circ}-\angle AFE = 180^{\circ}-\angle CFD = 180^{\circ}-\angle DAC = 150^{\circ}$