If the two circles ${(x-1)}^2+{(y-3)}^2=r^2$ and $x^2+y^2-8x+2y+8=0$ 

intersect in two distinct points; then

The given circles are ${(x-1)}^2+{(y-3)}^2=r^2$  and $x^2+y^2-8x+2y+8=0$ 

The centers are $\left(1,3\right),\left(4,-1\right)$

$C_1{C}_2=\sqrt{9+16}=5$

Since the two circles intersect at two distinct points$|r_1-r_2|<C_1{C}_2<r_1+r_2$

$\left|r-3\right|<5<r+3$

it implies that $r+3>5\Rightarrow r>2$

and $r-3<5\Rightarrow r<8$

Therefore, $2<r<8$