(2) x2-2x = (x-1) 2-1≥-1,y = (1/3) t is a decreasing function of t,

Therefore, y = (1/3) (x2-2x) ≤ (1/3) (-1) = 3,

It is also obvious that y > 0,

So the function value domain is (0,3).

(3) On (-∞, 1), t = x 2-2x is a decreasing function, and y = (1/3) t is a decreasing function about t,

Therefore, y = (1/3) (x 2-2x) monotonically increases on (-∞, 1);

On (1, +∞), t = x 2-2x is the increasing function, y = (1/3) t is the decreasing function of t,

Therefore, y = (1/3) (x 2-2x) monotonically decreases on (1, +∞).