Using suitable rearrangement, find the sum of the following:
$\frac{4}{7} + (-\frac{4}{9})+\frac{3}{7}+(-\frac{13}{9})$

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\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})

\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})

81^{\frac{1}{{log}_{5} 3}} + 27^{{log}_{9} 36} + 3^{\frac{4}{{log}_{7} 9}}

\frac{4}{7} + (\frac{- 4}{9}) + \frac{3}{7} + (\frac{- 13}{9})

The sum of the series $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + . . .$ to $\infty$ = $\frac{a}{e}$ . Find $(a + 3)^{2}$ .

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