Because ∠ ACF = 90, ∠ OAC+∠ ACF+∠ FCH = 180, ∠ OAC+∠ FCH = 90.

And ∠ OAC+∠ OCA = 90, ∠ FCH+∠ CFH = 90.

So ∠OAC =∠FCH∠ OCA =∠CFH, AC = CF

So △ OAC △ HCF, so FH=OC=m, similarly: DJ=OC=m=FH.

DJ∨FH, so ∠JDK=∠HFK, and ∠DKJ=∠FKH, so △ DKJ △ FKH.

So FK=KD, so k is the midpoint of FD.

3.g is the GM⊥x axis in M and E is the EN⊥x axis in N, which is the same as the second question: △ OAC △ MGA, △ OBC △ NEB.

So GM = ao = 2, ma = oc = m, bn = co = m, en = ob = 8, so G(-2-m, 2), E(8+m, 8).

So we can get q (3,5): f (-m, 2+m), D(m, m+8) from the second question.

So FQ= root number (2m 2+18) and DQ = root number (2m 2+18).

Then there is FQ=DQ, so DQ:FQ= 1, that is, ① is correct, and the radical (2m 2+18) is a function of m.

So it is not a fixed value, that is, ② is wrong, so two conclusions are correct and only one is correct.