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:

Circle 1: $C_1=(1,3)$, $R_1=r$.Circle 2: $x^2+y^2-8x+2y+8=0$.$C_2=(4,-1)$. $R_2=\sqrt{4^2+(-1{)}^2-8}=\sqrt{16+1-8}=\sqrt{9}=3$.Distance between centers $d=\sqrt{(4-1{)}^2+(-1-3{)}^2}=\sqrt{3^2+(-4{)}^2}=5$.For distinct intersection: $|R_1-R_2|<d<R_1+R_2$.
$$|r-3|<5<r+3$$
From $5<r+3⟹r>2$.From $|r-3|<5⟹-5<r-3<5⟹-2<r<8$.Since $r$ is a radius, $r>0$. Combined condition: $2<r<8$.