a) Let $AD\cap (I)=G$ so we have that $FGED$ is harmonic quadrilateral.Since $\overline{MN}\parallel\overline{BC}$ $$\Rightarrow -1=D(F,E,G,D)\overset{NM}{=}D(N,M,A,\infty_{NM})=(N,M,A,\infty_{NM})$$from that we get $A$ is the midpoint of $MN$ Hence $\overline{MN}\parallel\overline{BC}$ and $BD=BD$ we have $$\angle NFA=\angle BFD=\angle BDF=\angle ANF\Rightarrow AN=AF$$Similarly we have $AM=AE$ and hence $AE=AF$ so $NFEM$ is cyclic quadrilateral with center $A$ and radius $AM$. So we can easily get $K$ is the orthocentre of $\bigtriangleup DMN$ b) This is well known lemma, we can let the center of $(DMN)=O_1$ and draw the diameter $DO_1\cap (DMN)=D_1$ to prove this by $ND_1MK$ is a parallelogram , $D_1K\bot DL$ and $KL\bot DL$ .

Since $MAE$ is similar to $DCE$ and $NAF$ is similar to $DBF$, we get that $AM=AE=AF=AN$, and it is well known that that means $E$ and $F$ are the feet of the perpendiculars from $N$ to $DM$ and $M$ to $DN$ respectively. Since $A$ is the midpoint of $MN$, $K$ is the orthocenter of $DMN$, and $L$ is the intersection of $(DMN)$ and $(DEF)$, it is well known that these are collinear.