Good solutions here came from Sam, Dave,
Richard and Joe, Matthew and Ross, all at Madras College, St
Andrews. Other people tested special cases but did not prove
the general results.
If a 2 digit number has its digits reversed and the smaller of
the two numbers is subtracted from the larger we prove that
this difference can never be prime.
Let the 2 digit number be $a$ where $a> b$. Then $$ab - ba =
(10a + b) - (10b + a) = 9(a - b).$$
As $9(a - b)$ is a multiple of $9$, it is not prime.
Now let the 3 digit number be $abc$ $$abc - cba = (100a + 10b +
c) - (100c + 10b + a) = 99 (a - c).$$ As $99(a - c)$ is a
multiple of $99$, it is not prime.
The 4 digit number can be taken as $abcd$. $$abcd - dcba =
(1000a + 100b + 10c + d) - (1000d + 100c + 10b + a) = 9(111a +
10b - 10c - 111d).$$ Again, for any 4 digit number, this
difference is a multiple of 9 and so it can't be a prime
number.
Similarly for 5 digit numbers: $$\eqalign { abcde - edcba
&= (10,000a + 1000b + 100c +10d + e) - (10,000e + 1000d +
100c + 10b + a) \cr &= 99(101a + 10b - 10d -
101e).}$$
This number is a multiple of 99 so it will never be
prime.