Answer from cs61c-cj (Peter Lau 16539384) for Question 1
Since the cache has blocks 8 words in size, there are 32 entries in a block that will need to be addressed.  5 bits are necessary to represent 32 different offsets (32 = 2^5).  

53 (base 10) = 110101 (base 2)

The offset in the block is the 5 least significant bits or 10101.  Assuming there is need for an index field, its least significant digit is 1 and spans the next y bits of the memory (which are all 0).  So the index is just 1.  The tag field is the remaining most significant digits, which in this case is 0.