Answer from cs61c-bg (Spencer Ahrens 16187781) for Question 1
53 = 32'b000000000000000000000000000111001

8 words = 32 bytes = 2^5, so five bits in offset.  if it is 16KB, then 2^14/2^5 = 2^9, so 9 bits in the index.  This puts the contents at this memory address at the 27th byte of the second block of the chache with a remaining tag of zero