n is integer from 1-99

1.
If n is between 1-99,
probability that n is divisible by 3 = 1/3 (every 3 rd number is divisible by 3)
probability that n+1 is divisible by 3 = 1/3

Divisibility of n and (n+1) are mutually exclusive. n(n+1) is divisible by 3, if either n or (n+1) is divisible by 3.

P[n(n+1)] = P[n divisible by 3 OR (n+1) divisible by 3] = P[n] + P[(n+1)] = 1/3 + 1/3 = 2/3

2. Probabilty that n(n+1)(n+2) is divisible by 6 - Product of 3 consecutive numbers will be always divisible by 6. Probability = 1

3. Probabilty that n(n+1)(n+2) is divisible by 8

n is even:
(n+1) doesn't contribute to 8 divisibility, as this will be an odd number.

n(n+1)(n+2) - will be divisible by 8, if n(n+2) is divisible by 8.

n(n+2) will be divisible by 8, if one of them is divisible by 4. (8=4*2, Only one number can be divisible by 4, and we can get factor 2 from the other number)

for n between 1-99

P[n div. by 4] = 24/99 [had it been 1-100, we cud have got 1/4 - every 4th number)
P[(n+2) div. by 4] = 25/99 [from 3-101]
P[n(n+2) div. by 4 or 8] = 24/99 + 25/99 = 49/99

n is even, P[n(n+1)(n+2) div. by 8] = 49/99

n is Odd:
n(n+1)(n+2) - will be divisible by 8, if n+1 is divisible by 8.
for n between 1-99, n+1 is between 2-100.

between 2-100, we have 12 multiples of 8.
P[(n+1) div. 8] = 12/99

n is odd, P[n(n+1)(n+2) div. 8] = 12/99

Summing up even and odd cases,

Probabilty that n(n+1)(n+2) is divisible by 8 = 49/99 + 12/99 = 61/99

I think the answer for 1 and 2 have been discussed enough. I will add a bit on answer to problem no. 3 (similar to the method given by ConnectTheDots)

n can take 99 values from 1 to 99. n can be even or odd and depending on whether it is even or odd, we will handle the case.

Case 1: n is even.

n can take 49 different even values starting from 2, 4, 6 to 98.

In each case, (n+1) will be odd and (n + 2) will be even. Note that whenever you have 2 consecutive even numbers e.g. 2 and 4 or 6 and 8 or 96 and 98, exactly one of them will be a multiple of 4 (since you have a multiple of 4 in every 4 consecutive numbers (2 odd, 2 even)). So for all 49 values of even n, the given product will be a factor of 8.

Case 2: n is odd
Then only (n+1) is even. (n+1) must be a multiple of 8. There are only 12 multiples of 8 in the first 99 numbers (8, 16. ... till 96). So for only 12 values of odd n will the product be a multiple of 8.

Total there are 49 + 12 = 61 values of n for which the product will be a multiple of n.
Probability = 61/99

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