The equations of the two tangents from \((-5,-4)\) to the circle \(x^{2}+y^{2}+4 x+6 y+8=0\) are

Any line through the point \((-5,-4)\) is
\[
\begin{aligned}
y+4 &=m(x+5) \\
m x-y &+(5 m-4)=0
\end{aligned}
\]
Now, radius of circle
\[
=\sqrt{(2)^{2}+(3)^{2}-8}=\sqrt{4+9-8}=\sqrt{5}
\]
If it is a tangent, then perpendicular from centre \((-2,-3)\) is equal to the above radius.
\(\therefore \quad \frac{m(-2)-(-3)+(5 m-4)}{\sqrt{m^{2}+1}}=\sqrt{5}\)
\(\Rightarrow \quad-2 m+3+5 m-4=\sqrt{5} \sqrt{1+m^{2}}\)
\(\Rightarrow \quad 3 m-1=\sqrt{5} \sqrt{1+m^{2}}\)
\(\Rightarrow \quad(3 m-1)^{2}=5\left(1+m^{2}\right)\)
\(\Rightarrow \quad 9 m^{2}+1-6 m=5+5 m^{2}\)
\(\Rightarrow \quad 4 m^{2}-6 m-4=0\)
\(\Rightarrow \quad 4 m^{2}-8 m+2 m-4=0\)
\(\Rightarrow \quad 4 m(m-2)+2(m-2)=0\)
\(\Rightarrow \quad(m-2)(4 m+2)=0\)
\(\Rightarrow \quad \mathrm{m}=2,-\frac{1}{2}\)
Putting the value of \(m=2\) in Eq. (i), we get
\[
\begin{aligned}
2 x-y+5 \times 2-4 &=0 \\
\Rightarrow \quad 2 x-y+6 &=0
\end{aligned}
\]
Again, putting the value of \(\mathrm{m}=-\frac{1}{2}\) in Eq. (i), we get
\(-\frac{1}{2} x-y+5\left(-\frac{1}{2}\right)-4=0\)
\(\Rightarrow \quad-x-2 y-5-8=0\)
\(\Rightarrow \quad x+2 y+13=0\)