The question as stated is not quite answerable. A word has been defined to be 32-bits. We need to know whether the system is "byte-addressable" (you can access an 8-bit chunk of data) or "word-addressable" (smallest accessible chunk is 32-bits) or even "half-word addressable" (the smallest chunk of data you can access is 16-bits.) You need to know this to know what the lowest-order bit of an address is telling you.
Then you work from the bottom up. Let's assume the system is byte addressable.
Then each cache block contains 8 words*(4 bytes/word)=32=25 bytes, so the offset is 5 bits.
The index for a direct mapped cache is the number of blocks in the cache (12 bits in this case, because 212=4096.)
Then the tag is all the bits that are left, as you have indicated.
As the cache gets more associative but stays the same size there are fewer index bits and more tag bits.