Answer
The required value is $\frac{2\left( x+3 \right)}{\left( x+1 \right)}$ , $x\ne -1,-4,3,-3$.
Work Step by Step
Consider the expression,
$\frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}$
The given expression can be further evaluated by inverting the divisor and multiplying:
$\begin{align}
& \frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}=\frac{2x+8}{x-3}\cdot \frac{{{x}^{2}}-9}{{{x}^{2}}+5x+4} \\
& =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{{{x}^{2}}+4x+x+4} \\
& =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{x\left( x+4 \right)+1\left( x+4 \right)} \\
& =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{\left( x+4 \right)\left( x+1 \right)}
\end{align}$
Solve further:
$\begin{align}
& \frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}=\frac{2\left( x+3 \right)\left( x+4 \right)\left( x-3 \right)}{\left( x+1 \right)\left( x+4 \right)\left( x-3 \right)} \\
& =\frac{2\left( x+3 \right)}{\left( x+1 \right)}
\end{align}$
Here, $x\ne -1,-4,3,-3$ because at $x=-1,-4,3,-3$ the denominator will become zero.
Therefore, the value of $\frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}$ is $\frac{2\left( x+3 \right)}{\left( x+1 \right)}$.