3 & 4 card hands

I am trying to find out the method (and the answer) to the following:

What is the probability of getting a 3 card straight and a 3 card straight flush, or a 4 card straight and a 4 card straight flush in a FIVE CARD POKER HAND.
I have some knowledge of probabilities, but this is beyond my current abilities.
Thanks in advance.

One way is to break the hands down into various types, and then see how many of each are straights and SFs.
For instance think
(i) No pair - five different cards. Some combinations will have
(a) No possible straights (e.g. A K J T 8)
(b) Others will have one (e.g. A K Q T 9)
(c) Others will have two (e.g. A K Q J 9, A K Q 3 2)
(d) Others will have three (e.g. A K Q J T),
(ii) One pair - hence four different cards where one of them is in two suits
(iii) Two pairs - three different cards where two of them are in two suits...
etc.
This might help break it down into manageable chunks.

When you say 3-card straight, does this include a hand that has two pair, or three of a kind?
For example, 6s 7h 7c 8s 8d, or Jd Qc Ks Kh Kc?

The hardest part might be trying to count the 3-card straights that aren't also 4-card straights.

Let's use straight flushes as an example, as that's easier to do, since there aren't that many ways to count a hand multiple times.
How many 4-card straight flushes are there?
For each one, how many cards are there that can be the fifth card?
I assume you do not want to count a fifth card that makes a 5-card straight flush.
Example: with 2-5 of spades, there are 48 cards remaining, but you skip the ace and 6 of spades as those form a 5-card straight flush. This is 46 hands. However, some of these hands are pairs (e.g. 4 of diamonds), 5-card straights (e.g. ace of clubs), and 5-card flushes (e.g. 9 of spades); do you want to count those as 4-card straight flushes?

^ Yes that's an alternative way to look at the problem and you can see whether it comes up with the same answer (I always like to have the same answer using two different ways).
(i) 5-card straights (AKQJT...5432A)
(ii) 4-card straights (AKQJx, x432A, KQJTx...x5432)
(iii) 3-card straights (AKQ32, AKQxx...xx321)
Actually if you're only looking at non-pair hands the five card hands are probably easy to generate (card1 = A to 6, card2 = card1+1 to 5, etc.)

Yes you must always calculate

5-card straights
4-card straights (minus 5-card straights)
3-card straights (minus 4,5-card straights)

If you don't do that, then you will calculate a 5-card straight hand to be three 3-card straight hands and a 4-card straight hand to be two 3-card straight hands.

After you have calculated the frequency of 5 card straights, you must calculate:

Given a hand with 5 different ranks: what is the probability of having a

5-card flush
4-card flush (minus 5-card flushes)
3-card flush (minus 4,5-card flushes)

You can combine those flush probabilities (for hands with 5 ranks) with the frequencies of various 3,4,5-card straights to then calculate anything you want.