$1 . \underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots 7}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} 8$ divided by 72 has a remainder of
$$
\begin{array}{l}
\underbrace{66 \cdots 6}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{88 \cdots 8}_{100 \uparrow} \\
=\underbrace{66 \cdots 677 \cdots}_{10 \uparrow} \underbrace{77}_{100 \uparrow} \underbrace{88 \cdots 82000}_{96 \uparrow}+72 \times 95+48 . \\
\underbrace{66 \cdots}_{100 \uparrow} \underbrace{77 \cdots}_{100 \uparrow} \underbrace{78 \cdots}_{96} 82000=a .
\end{array}
$$

Since the sum of the digits of $a$
$$
100 \times 6 + 100 \times 7 + 96 \times 8 + 2 = 2070
$$

is divisible by 9, thus $a$ is divisible by 9; also, the last three digits of $a$ are divisible by 8, so $a$ is also divisible by 8. Since 8 and 9 are coprime, therefore, $a$ is divisible by $8 \times 9 = 72$. Hence, the remainder of the original number when divided by 72 is 48.