A humble try for Q1....
CASE I : When (x-y) is divisible by 5
The following pairs of integers from the set when subtracted give 5 :
(6, 1) , (1, 6)
(7, 2), (2, 7)
(8, 3), (3, 8)
(9, 4), (4, 9)
(10, 5), (5, 10)
(11, 6), (6, 11)
(12, 7), (7, 12)
(13, 8), (8, 13)
(14, 9), (9, 14)
(15, 10), (10, 15)
And for 10,
(15, 5), (5, 15)
Also there are cases where the end digit can be zero. There would be 14 such cases.
Probability : Number of Pairs/Subtractions Possible
= 35/214 (There are 14 numbers in the set)
CASE II : When (x+y) is divisible by 5
The following pairs of integers when added yield 5 :
(1, 4), (4, 1)
(2, 3), (3, 2)
And there are no more.
But for further multiples of 5, such as 10,
(1, 9), (9, 1)
(2, 8), (8, 2)
(3, 7), (7, 3)
(4, 6), (6, 4)
And similarly for 15,
(1, 14), (14, 1)
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(8, 7), (7, 8)
Probability : 28/214
Needless to say either the first case or the second case must hold for overall divisibility by 5.
Adding the two,
Probability : 63/214
Am not at all sure of this answer. Just a try.