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Solution:

I will analyze this step by step:

$e ^{\ln {4}} = e ^{\log _{e} {4}} = 4$

$\log _{12321}{1024} = \log _{111}{32}$ therefore: $\frac{\log _{111} {1024}}{\log _{12321}{1024}} = \frac{\log _{111} {1024}}{\log _{111}{32}} = \log _{32}{1024} = 2$

$2 \times \log _{9} {0.\dot{3}} = 2 \times \log _{9} { \frac{1}{3} } = 2 \times \log _{9} ({\frac{1}{9}})^{1/2} = 2 \times (- \frac{1}{2}) = -1$

So, the first part is actually $x > 4 + 2 - 1$ or just $\boxed{x > 5}$

$\log _{2} {512^{34}} = 34 \times \log _{2} {512} = 34 \times 9 = 306$

$\frac{\log _{49} {25^{4096}} }{\log _{\sqrt{7}} {5}} = \frac{4096 \times \log _{49} {25} }{\log _{7} {25}} = \frac{4096 \times \log _{49} {25} }{\log _{49} {25 ^{2}}} = \frac{4096 \times \log _{49} {25} }{2 \times \log _{49} {25}} = 2048$

$4086 \times \log _{25}{5} = 4086 \times \frac{1}{2} = 2043$

So, the second part is $x < 306 + 2048 - 2043$ or just $\boxed{x < 311}$

When we unite the first and second part, we get $5 < x < 311$ , so solution for equation is $\boxed{x \in (5, 311)}$ .

$311 - 5 = \boxed{306}$

\begin{aligned} e^{\ln 4} + \frac{\log_{111} 1024}{\log_{12321}1024} + 2\log_9 0.\dot{3} < & x < \log_2 512^{34} + \frac{\log_{49}25^{4096}}{\log_\sqrt{7} 5} - 4086\log_{25} 5 \\ 4 + + \frac{\log_{111} 1024}{\frac{\log_{111} 1024}{\log_{111} 12321}} + 2\log_9 \frac{1}{3} < & x < \log_2 2^{9 \times 34} + \frac{\log_\sqrt{7} 5^{2\times 4096}}{\log_\sqrt{7} 49 \log_\sqrt{7} 5} - 4086\log_{25} 25^\frac{1}{2} \\ 4 + \log_{111} 111^2 - 2 \log_9 9^\frac{1}{2} < & x < 306 + \frac{8192}{4} - 2043 \\ 4 + 2 - 1 < & x < 306 + 2048 - 2043 \\ 5 < & x < 311 \\ & x \in (5, 311) \\ \Rightarrow n-m & = \boxed{306} \end{aligned}