Solution:

44 and 66

Solution:

Just 1

Solution:

The digits are 12, 5, 6, and 8

Solution:

3211000

Solution:

129.

These are the first 10 prime numbers (2, 3, 5...) prefixed with a 1

Solution:

20 times

8,18,28,38,48,58,68,78,98,80,81,82,83,84,85,86,87,88,88(2),89

Solution:

Starting with 1 marble in a bag, each bag contains double the number of marbles as the previous bag.
ie 1,2,4,8,16,32,64,128,256,512.

That way whatever number of marbles I would ask for you would just add up the boxes to equal the number asked.

Solution:

1) 8/(3-(8/3))
= 8/(1/3)
= 24

2)((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24

3)(3!/)*8

4)(8-3)!/(8-3)
( × )!
( + )!
?(8×8×3×3)
8+(8×(3!/3))
((?(8+8) × (3/3))!
?(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!

Solution:

9376

Looking at the last digit, the last digit must be either 0, 1, 5 or 6.
Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76.
Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376.
Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376.
Out of those, only 9376 is a 4 digit number

Solution:

2519

The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.

Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.