The circle $x^2+y^2=4x+8y+5$ intersects the line $3x-4y=m$ at two distinct points if:

Circle equation: $x^2-4x+y^2-8y-5=0$.Center $C(2,4)$.Radius $r=\sqrt{(-2{)}^2+(-4{)}^2-(-5)}=\sqrt{4+16+5}=5$.A line intersects a circle at two distinct points if the perpendicular distance $p$ from the center to the line is less than the radius.Line: $3x-4y-m=0$.$p=\frac{|3(2)-4(4)-m|}{\sqrt{3^2+(-4{)}^2}}=\frac{|6-16-m|}{5}=\frac{|-10-m|}{5}=\frac{|10+m|}{5}$.Condition: $p<r⟹\frac{|10+m|}{5}<5$.$|10+m|<25$$-25<10+m<25$$-35<m<15$.