Welcome to the forums! :)
There's actually a Mathematics 2U Board you might want to check out a bit later (to post further questions, and to view other questions that you may have that have already been asked! otherwise there are more resources there :) )
For part i), consider this: in what cases do you have no chance of winning before the game starts? Essentially, this maps out a bit like a game of bullshit; if you roll a certain number on the first die, and you roll one up, one down or the same on the second, you instantly have no chance of winning. Drawing a 6x6 table with each outcome helps visualise this best, but you can also list them out case by case :) Try doing this yourself, and then check the spoiler for the solution :)
Spoiler
(1, 1), (1, 2), (2, 1), (2, 2), (2, 3) ... (6, 6) = 16 cases out of a possible 36, ie 4/9
ii) This is a little trickier. There are a few different ways we can consider this;
a) By the number of numbers in between each pair
b) case-by-case (which takes a lot longer)
a) is the easiest option, but you can have a look at b) if you have a lot of time.
You can have anywhere from 1-4 numbers in between each pair ie. for 1, an example (2, 4) and for 4, an example would be (1, 6).
There are eight cases with 1 number in between: hence, we multiply 8/36 (the chance of getting this case) by the chance of hitting a number in between on the third die, which is 1/6.
There are six cases with 2 numbers in between: repeating, we have 6/36 x 2/6
There are four cases with 3 numbers in between: repeating, we have 4/36 x 3/6
There is one case with 4 numbers in between: and lastly we have 2/36 x 4/6
Adding all these probabilities will give you the answer! :)
Also, if I recall correctly, there was a similar question in the 2018 HSC, you might want to check that out! (unless this is from the 2018 HSC?). If you have any further questions, please enquire further! :)
Hope this helps!!