Here is the code for part (a) of the problem:
ldi R19, 0 ; set R19 to zero
ldi R20, $15 ; load $15 into R20
sts $67, R20 ; store $15 in memory location $67
add R19, R20 ; add $15 to R19
subi R24, 1 ; decrement counter
brne loop ; loop until counter is zero
And here is the code for part (b) of the problem:
ldi R19, 0 ; set R19 to zero
ldi R20, $15 ; load $15 into R20
sts $67, R20 ; store $15 in memory location $67
add R19, R20 ; add $15 to R19
subi R24, 1 ; decrement counter
brne loop ; loop until counter is zero
sts $70, R19 ; store result in memory location $70
Note that in both cases, we use a loop to add $15 to R19 seven times. The loop is controlled by a counter in R24, which is initialized to 7 and decremented by 1 in each iteration of the loop. The brne instruction branches to the loop label if the counter is not zero. Finally, in part (b), we store the result in memory location $70 using the sts instruction.