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

$\begin{array}{l}{x}^2+y^2=4x+8y+5\\ x^2+y^2-4x-8y-5=0\end{array}$

Centre $g,f=(+2,+4)$

Radius $=\sqrt{g^2+f^2-c}=\sqrt{4+16+5}=5$

Given lines touches circle at two distinct points.  So perpendicular distance from centre to line is less than the radius

$\begin{array}{l}\left|\frac{3(+2)-4(+4)-m}{\sqrt{3^2+4^2}}\right|<5\\ \left|\frac{+6-16-m}{5}\right|<5\Rightarrow \left|\frac{-10-m}{5}\right|<5\\ -5<\frac{10+m}{5}<5\\ -25<10+m<25\\ =-35<m<15\end{array}$