The solution solves for the factors and prime factors of given numbers. It seems to be a two-part question.

• Factors of 8: $1, 2, 4, 8$
• Factors of 12: $1, 2, 3, 4, 6, 12$
• Factors of 20: $1, 2, 4, 5, 10, 20$ The factors are: 8: $\boxed{\text{1, 2, 4, 8}}$, 12: $\boxed{\text{1, 2, 3, 4, 6, 12}}$, 20: $\boxed{\text{1, 2, 4, 5, 10, 20}}$.

The common factors of 8, 12, and 20 are $1, 2, 4$.
The common factors are $\boxed{\text{1, 2, 4}}$.

The greatest common factor is 4.
The greatest number is $\boxed{\text{4}}$.

The prime factors are: 48: \boxed{\text{2^4 \times 3^1}}, 60: \boxed{\text{2^2 \times 3^1 \times 5^1}}.

Common prime factors are 2 and 3. The lowest power of 2 is $2^2$, and the lowest power of 3 is $3^1$.

The product of the common prime factors is $\boxed{\text{12}}$.

Yes, the product (12) is the H.C.F. of 48 and 60.
Reason: The H.C.F. is found by taking the product of the common prime factors, each raised to the lowest power that appears in the prime factorization of the given numbers.
The product is $\boxed{\text{Yes}}$, because it is the product of the common prime factors raised to their lowest powers.

The common prime factors are 3 and 5. The lowest power of 3 is $3^1$, and the lowest power of 5 is $5^1$.

The greatest number that divides these three numbers exactly is $\boxed{\text{15}}$.