1st april fool

5.(5)⁹⁸
=5.(25)⁴⁹
=5.(26-1)⁴⁹.

expand binomially,
5.[ ⁴⁹C₀(26)⁴⁹ - ⁴⁹C₁(26)⁴⁸.............. +⁴⁹C₄₈(26) - ⁴⁹C₄₉]
26 is a factor of 13.
therefore the remainder is..
13-5 = 8. :)

@joydoot : nice solution :)

Here is my solution using elementary and pure number theory

5² ≡ 25 mod 13
≡ -1 mod 13

(5²)⁴⁹ ≡ (-1)⁴⁹ ≡ -1 mod 13...............1

and 5 = 5 mod 13................2

multiplying 1 and 2

5⁹⁹ ≡ -5 ≡ 8 mod 13

:)