$7.68 \log _{a} \sqrt{4+x}+3 \log _{a^{2}}(4-x)-\log _{a^{4}}\left(16-x^{2}\right)^{2}=2$.

For what values of $a$ does the equation have a solution?
7.68 Given the domain of the logarithmic function, the square root, and the restrictions imposed on the base of the logarithm, we conclude that

$$
\left\{\begin{array} { l } 
{ 4 + x > 0 , } \\
{ 4 - x > 0 , } \\
{ a > 0 , a \neq 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
-4 < x < 4, \\
a > 0, a \neq 1
\end{array}\right.\right.
$$

We will convert all terms of the equation to the base \(a^{2}\): \(\log _{a^{2}}(4+x)+3 \log _{a^{2}}(4-x)-\log _{a^{2}}\left(16-x^{2}\right)=2\), or \(\log _{a^{2}} \frac{(4+x)(4-x)^{3}}{(4-x)(4+x)}=2\).

By canceling \((4-x)(4+x)\) in the fraction (there will be no loss of roots since \((4-x)(4+x)>0\) under condition (1)), we get \(\log _{a^{2}}(4-x)^{2}=2\), from which \(2 \log _{a^{2}}(4-x)=2\), since \(4-x>0\). Therefore, \(x=4-a^{2}\).

We will find those values of \(a\) for which \(-4 < x < 4\), \(a > 0\), and \(a \neq 1\). Substituting \(x=4-a^{2}\) into the inequality \(-4 < 4-a^{2} < 4\), we get \(0 < a^{2} < 8\). Since \(a > 0\) and \(a \neq 1\), we obtain \(0 < a < 1\) and \(1 < a < 2 \sqrt{2}\).

Answer: \(x=4-a^{2}\), where \(0 < a < 1\) and \(1 < a < 2 \sqrt{2}\).