Solution:

(1) Set A can be written in roster form as:

A = {1, 3, 9, 27, 81, ...}

Explanation:

- The set A is defined as the set of all terms an, where n is a natural number and an is such that an = 3an-1 and a1 = 1.
- The first term a1 is given as 1.
- To find the second term a2, we use the formula an = 3an-1. Thus, a2 = 3a1 = 3.
- Continuing in this way, we can find the third term a3, fourth term a4, and so on.
- The terms in the set A are 1, 3, 9, 27, 81, and so on, where each term is three times the previous term.

(2) Set B can be written in roster form as:

B = {1, 1, 2, 4, 8, 16, 32, ...}

Explanation:

- The set B is defined as the set of all terms an, where n is a natural number and an is such that an^2 = an-1an and a1 = a2 = 1.
- The first two terms a1 and a2 are given as 1.
- To find the third term a3, we use the formula an^2 = an-1an. Thus, a3^2 = a2a1 = 1, which gives us two possibilities for a3: a3 = 1 or a3 = -1. However, a3 cannot be negative, so we take a3 = 1.
- Continuing in this way, we can find the fourth term a4, fifth term a5, and so on.
- The terms in the set B are 1, 1, 2, 4, 8, 16, 32, and so on, where each term is the product of the two previous terms.

Note: The set B is also known as the Fibonacci sequence, where each term is the sum of the two previous terms. However, this is not the case here as the terms in set B are the product of the two previous terms.