EDIT: Sorry I didn't enforce that both digits must appear in the part below, so it's wrong. I am too busy now to correct it.
When you fix the digits it's then 2^5 (think why)
So overall you get C(10,2) * 2^5
(2) You can use inclusion exclusion. And you can rephrase question as asking how many ways to deal a five card hand such that at most two suits appear.
Number of ways to select two suits from four is C(4,2)
When two suits are fixed, number of ways to select 5 cards is C(26,5)
Note that for each suit, we count the number of ways to select 5 cards with just that suit 3 times.
For example, hands with all hearts cards are included in (hearts and spades), (hearts and clubs), and (hearts and diamonds)
Thus we should subtract the duplicates, so finally we get
C(4,2) * C(26,5) - 4 * 2 * C(13,5)