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Let's address each of these problems one by one:

1. How many positive integers less than 2000 are of the form x^n for some positive integer x and n >= 2?

To solve this problem, we need to count the number of positive integers that can be expressed in the form x^n, where x is a positive integer and n is greater than or equal to 2.

Let's start with the smallest possible value of x, which is 2. In this case, we can have values like 2^2, 2^3, 2^4, ... up to the largest value of n such that 2^n is less than 2000. Similarly, we can do this for all values of x from 2 to the largest value of x such that x^n is less than 2000.

For x = 2:
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256 (which is greater than 2000, so we stop here)

For x = 3:
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243 (which is greater than 2000, so we stop here)

Continue this process for x = 4, 5, 6, 7, ... until you reach the largest possible value of x such that x^n is less than 2000.

Now, count the unique positive integers you've found. The total count will be the answer to this problem.

2. Find the smallest positive integer n such that the product 1999n ends with 888.

To find the smallest positive integer n that satisfies this condition, you need to determine the remainder when the product 1999n is divided by 1000.

1999n ≡ 888 (mod 1000)

Now, you can solve for n using modular arithmetic.

Let's find the modular inverse of 1999 modulo 1000:
1999⁻¹ ≡ 999 (mod 1000)

Now, multiply both sides of the congruence by 999 to isolate n:
n ≡ 888 * 999 ≡ 444 * 3 * 111 * 9 ≡ 333 * 111 * 9 ≡ 333 * 9 ≡ 999 (mod 1000)

So, the smallest positive integer n that satisfies the given condition is n = 999.

3. Replace each of the boxes with a distinct number from 1, 2, 3, 4, 5, 6, 7, 8, 9 so that the sum is as close as possible to 1 without being equal to or greater than 1. What is the sum expressed as a common fraction?

This is a classic problem in mathematics. You want to distribute the numbers 1 through 9 into the boxes in such a way that the sum is as close as possible to, but less than 1.

One way to approach this problem is to use trial and error. You can start by trying different combinations of numbers in the boxes until you find a combination that satisfies the condition.

Here's one possible solution:
Box 1: 1/9
Box 2: 2/9
Box 3: 3/9 (which simplifies to 1/3)
Box 4: 4/9

The sum of these fractions is:
1/9 + 2/9 + 1/3 + 4/9 = 8/9

So, the sum is 8/9, which is as close as possible to 1 without exceeding it.