Three. (50 points) A function calculator has a display screen and two operation keys. If the first operation key is pressed once, the number on the display will change to $\left[\frac{x}{2}\right]$ (where $[x]$ represents the greatest integer not exceeding the real number $x$); if the second operation key is pressed once, the number on the display will change to $4x+1$. Pressing either operation key once is called one operation. The number on the display is currently 1. Answer the following questions:
(1) Can the number 2000 appear on the display after a finite number of operations? Explain your reasoning.
(2) How many integers less than 2000 can appear on the display after a finite number of operations?
Three, (1) Impossible.
Convert the number to binary. Then pressing the first operation key means removing the last digit of the number on the display; pressing the second operation key means appending 01 to the number on the display.

When the initial number is 1, after performing the above two operations, the resulting number does not have two 1s adjacent. However, 2000 in binary is $(11111010000)_2$, which has two 1s adjacent. Therefore, the number 2000 cannot appear on the display.
(2) If the second operation is performed first, followed by the first operation, it is equivalent to appending a 0 to the original number. Hence, any binary number without two 1s adjacent can be obtained after a finite number of operations.

Thus, the number of integers less than 2000 that can appear on the display is equivalent to the number of natural numbers not greater than $(10101010101)_2$ and without two 1s adjacent.

Moreover, the number of natural numbers with $k(0 \leqslant k \leqslant 6)$ 1s and not greater than $(10101010101)_2$ is $\mathrm{C}_{12-k}^{k}$. Therefore, the number of natural numbers that meet the condition is
$$
1+11+45+84+70+21+1=233 .
$$