Solution:

1. Addressing the main memory:
Since the main memory contains blocks, and each block contains 8 bit words, we need to find the total number of blocks in the main memory.

Let the total number of blocks in the main memory be 'N'.
Since the block-set associative cache memory consists of blocks divided into four block sets, there are N/4 blocks per set.
Since there are four block sets, the total number of sets in the cache is N/4 * 4 = N.

Let the number of bits required to address the main memory be 'b'.
Each block contains 8 bit words, so the address of each word within a block can be represented using 3 bits (since 2^3 = 8).
Therefore, each block can be addressed using log2(8) = 3 bits.
Since there are N blocks in the main memory, we need log2(N) bits to address the main memory.

Thus, we have b = log2(N) + 3.

Substituting N = 2^16 (since each block is 16 bits long), we get:
b = log2(2^16) + 3
b = 16 + 3
b = 19 bits.

However, we need to calculate the number of bits required to address the main memory in the block-set associative cache memory, which is divided into four block sets.
Therefore, we need to find the number of bits required to address each set in the cache, which is given by:
log2(N/4) + 3 = log2(2^14) + 3 = 14 + 3 = 17 bits.

Therefore, the total number of bits required to address the main memory in the block-set associative cache memory is:
b = log2(N) - log2(4) + log2(4) + 3 = log2(N/4) + 2 + 17 = 22 bits.

2. Representing the TAG, SET, and WORD fields:
Since the cache memory is block-set associative, each block in the main memory maps to one of the four sets in the cache.
Let the number of bits required to represent the TAG, SET, and WORD fields be 't', 's', and 'w', respectively.

Since there are four sets in the cache, we need log2(4) = 2 bits to represent the SET field.
Since each block is 16 bits long, we need log2(16) = 4 bits to represent the WORD field.

To calculate the number of bits required to represent the TAG field, we can subtract the number of bits used to represent the SET and WORD fields from the total number of bits required to address the main memory, which is 22 bits.
Therefore, we have:
t = 22 - s - w
t = 22 - 2 - 4
t = 16 bits.

Thus, the number of bits required to represent the TAG, SET, and WORD fields are:
TAG = 16 bits.
SET = 2 bits.
WORD = 4 bits.