∴△=(-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.