∴△=(-3)2-4(k- 1)≥0,
The solution is k≤ 134,
∫k is a positive integer,
The value of ∴k is 1, 2, 3;
(2) When k= 1 and x2-3x=0,
Obviously, the root of this equation is 0,
When k=2, x2-3x+ 1=0,
△=(-3)2-4(2- 1)=5,
This equation has two unequal real roots,
When k=3, x2-3x+2=0,
The solution is x 1= 1, x2=2,
The equation has two integer roots that are not zero.
∴k=3,
The quadratic function is y=x2-3x+2,
∫ The secondary image is shifted down by 2 units,
∴ The analytical formula of the translation function image is y = x2-3x;
(3) When the straight line y=5x+b passes through the point (0,0), b=0,
And y = 5x+by = x2? 3x,
If y is removed, x2-8x-b=0,
When two function images intersect, △=(-8)2-4× 1×(-b)=0,
The solution is b=- 16,
Therefore, when the straight line y=5x+b has three common points with the image g, the value range of b is-16 < b < 0.