Answer from cs61c-en (Lai, Siu Lun Chris 16529329) for Question 2
From the ID/EX register, another mux, and the EX/MEM register.

ID/EX: add $2 $3 $4
       addi $5 $6 1

another:  add $5 $3 $4
mux       sll $2 $2 0
          sll $2 $2 0
          sll $2 $2 0
          add $7 $0 $5

EX/MEM:   add $5 $3 $4
          sll $2 $2 0
          sll $2 $2 0
          add $7 $0 $5