Analysis: y=kx-2k+6=k(x-2)+6, so it passed the fixed point (2,6).

(2) Solution:

M(6, y) can be set from (1).

tan∠QOM = tan(∠QOx-∠MOx)=(tan∠QOx-tan∠MOx)÷( 1+tan∠QOx×tan∠MOx)

=(6÷2-y÷6)÷( 1+6÷2×y÷6)= tan 45 = 1

Solution, y=3

∴M(6,3)

Formula (y-3)÷(x-6)=(3-6)÷(6-2) from two points.

The equation of line QM is 3x+4y-30=0.

(3)/* ∠AEO = 45, and the angle is certain, so point E can be regarded as a point on the arc, and ∠AEO is the circumferential angle of the arc, and ∠ AEO's central angle is 90 */

By the known A (6 6,0), B (0,6), C (0,3)

Let the midpoint N (3,0) of AO intersect with N as the vertical line L of AO, and take a point D above the first quadrant L, so that DN=AN, then ∠ ODA = 90, then E is on a circle with D (3,3) as the center and DA length of 3√2 as the radius in the first quadrant.

/* Take a point f on ⊙D and connect CF, CD and FD, then CF≤FD+CD. When f is on the extension line of CD, the equal sign holds */

Then when E is located on the extension line of CD, CE is the longest, and CE(max)=3+3√2.

I am a high school student. If the method doesn't suit you, you can ask. I didn't know how to do congruence and auxiliary lines, but I thought of adding them again.