Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$. Find the remainder when $x$ is divided by $1000$.
The first condition implies
\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]
\[128+a^{128} = \log_a\log_a 2^{24}\]
\[a^{a^{128}a^{a^{128}}} = 2^{24}\]
\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]
So $a^{a^{128}} = 8$. 
Putting each side to the power of $128$:
\[\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},\]
so $a^{128} = 64 \implies a = 2^{\frac{3}{64}}$. Specifically,
\[\log_a(x) = \frac{\log_2(x)}{\log_2(a)} = \frac{64}{3}\log_2(x)\]
so we have that
\[256 = \log_a(\log_a(x)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[12 = \log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[2^{12} = \frac{64}{3}\log_2(x)\]
\[192 = \log_2(x)\]
\[x = 2^{192}\]
We only wish to find $x\bmod 1000$. To do this, we note that $x\equiv 0\bmod 8$ and now, by the Chinese Remainder Theorem, wish only to find $x\bmod 125$. By [Euler's Totient Theorem](https://artofproblemsolving.com/wiki/index.php/Euler%27s_Totient_Theorem):
\[2^{\phi(125)} = 2^{100} \equiv 1\bmod 125\]
so
\[2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125\]
so we only need to find the inverse of $6 \bmod 125$. It is easy to realize that $6\cdot 21 = 126 \equiv 1\bmod 125$, so 
\[x\equiv 21\bmod 125, x\equiv 0\bmod 8.\]
Using [Chinese Remainder Theorem](https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem), we get that $x\equiv \boxed{896}\bmod 1000$, finishing the solution.