(1) since the 2 cards are drawn simultaneously, it is a case of "without replacement", meaning that you cannot have 2 cards with same number.

total outcome = 9x8 = 72
for sum of 5 or less, possible combinations are ...
12,13,14,21,23,31,32,41 (8 combinations)
P(sum 5 or less) = 8/72 = 1/9

if it is draw 1 card, replace, draw another card,
it would be "with replacement", then we can consider the combinations of 11 & 22
for "with replacement",
P(sum 5 or less) = 10/81
so cannot figure out how you can get 1/6

(2)(ii)
for "at least one draw", need to consider 3 cases ...
case 1: 5 or less on BOTH draws
case 2: 5 or less on 1st draw but not on 2nd
case 3: 5 or less on 2nd draw but not on 1st
P(5 on less on at least one draw)
= 1/9 x 1/9 + 1/9 x 8/9 + 8/9 x 1/9 = 17/81

alternatively, use complement method
P(5 or less on at least one draw)
= 1 - P(5 or less on 0 draw)
= 1 - 8/9 x 8/9 = 17/81

Answers :
A /B = 1/9
CD/EF = 29/12
G/HI = 1/81
JK/LM = 17/81

Thank you for your explanation, but I have one question regarding your answer. Why is the total no. if possible outcomes 9x8?
I thought since the order was not important I used the combination formula instead of the permutation formula.
I did 9!/[2!*(9-2)!] and got 36.

Also, how did you make sense of CD and EF?

if you consider order as not important, then you should do likewise for the possible combinations, meaning ...
12 is same as 21,
13 is same as 31
14 is same as 41
23 is same as 32
so you have 4 combinations
P(sum 5 or less) = 4/36 = 1/9

Thank you very much for your help.

for CD/EF

note that ...
sum 3 possible only with cards 1&2 ;
probability = 1/36
sum 4 possible only with cards 1&3 ;
probability = 1/36
sum 5 possible only with cards 1&4, and 2&3 ; probability = 2/36

summary ...
sum | score | probability
3 | 10-3 = 7 | 1/36
4 | 10-4 = 6 | 1/36
5 | 10-5 = 5 | 2/36
6 to 17 | 2 | 32/36

expected score
= summation(score x probability)
= 7 x 1/36 + 6 x 1/36 + 5 x 2/36 + 2 x 32/36
= 87/36
= 29/12