$-$ Segundo enunciado, podemos escrever: \begin{matrix} (h \circ f) &=& arctg (\frac{5+7^x}{4}) &=& \alpha \\ (h \circ g) &=& arctg (\frac{5-7^x}{4}) &=& \delta \end{matrix} $-$ Dado o comando, temos: \begin{matrix} (h \circ f)(a) &+& (h \circ f)(a) &=& \frac{\pi}{4} \\ \\ \alpha &+& \delta &=& 45^{\circ} \end{matrix} $\color{orangered}{Obs:}$ $\tan{(x+y)} = \frac{\tan{(x)} + \tan{(y)}}{1-\tan{(x)}.\tan{(y)}}$ \begin{matrix} \tan{(\alpha + \delta)} = \tan{45^{\circ}} \\ \\ {\Large{\frac{\tan{(\alpha)} + \tan{(\delta)}}{1-\tan{(\alpha)}.\tan{(\delta)}} }}= 1 \\ \\ \large{\frac{5+7^a}{4} + \frac{5-7^a}{4} = {\normalsize{1}} - \frac{5+7^a}{4} .\frac{5-7^a}{4} } \\ \\ 7^{2a} = 49 \\ \\ \fbox{$a = 1$} \end{matrix} Assim, \begin{matrix} f(a) - g(a) = \large{\frac{5+7^a}{4} - \frac{5-7^a}{4}} \\ \\ \fbox{$ f(a) - g(a) = \frac{7}{2}$} \\ \\ \\ Letra \ (D) \end{matrix}