II. Fill-in-the-blank Questions (9 points each, total 54 points)  
1. Remove all perfect squares and cubes from the natural numbers, and arrange the remaining numbers in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2008}=$ $\qquad$
Two, 1.2062.
First, remove numbers of the form $n^{6}$. In the intervals $\left(1^{6}, 2^{6}\right),\left(2^{6}, 3^{6}\right),\left(3^{6}, 4^{6}\right), \cdots$, there are no more overlaps between square numbers and cube numbers.

Since $\sqrt{2^{6}}=2^{3}=8, \sqrt[3]{2^{6}}=4$, removing the interval endpoints, we know that $\left(1^{6}, 2^{6}\right)$ contains 6 square numbers and 2 cube numbers. Similarly, $\left(2^{6}, 3^{6}\right)$ contains 18 square numbers and 4 cube numbers.

Therefore, in the interval $\left[1,3^{6}\right]$, there are a total of 33 square and cube numbers.

Since $2008+33>45^{2}$, and $45^{2}=2025,46^{2}=$ $2116,12^{3}=1728,13^{3}=2197$, the interval $(2025$, $2116)$ contains no square or cube numbers. In $\left(3^{6}, 45^{2}\right]$, there are 18 square numbers $\left(28^{2}, 29^{2}, \cdots, 45^{2}\right)$ and 3 cube numbers $\left(10^{3}, 11^{3}, 12^{3}\right)$. Thus, $2008+33+18+3=2062$, so $a_{2008}=2062$.