Answer from cs61c-aq (Anjali Koppal 16454392) for Question 2
This set of instructions would lead to a data hazard, because the same variable needs to be accessed in both instructions, so the second will have to stall till the first is done with using the instruction. we can avoid this by 'bypassing' or sending the result of the first instruction to the second with minimal delay. but since $t0 does not get its value till the very last stage, it seems inevitable that we will have to stall for a little while.