The last two digits of 7^(2008) are

7¹= 07
7²= 49
7³= 343
7⁴= 2401
7⁵= 16807
7⁶= 117649
7⁷= 823543
7⁸= 5764801
Here, note that the last twodigitsof 7⁴and 7⁸are 01. For every 4th power, the last digits are 01
Since 2008 is divisible by 4, the last digits of7²⁰⁰⁸are 01.
THanks

7 = 07 (mod 100) 7 ^ 2 = 49 (mod 100) 7 ^ 4 = 01 (mod 100) So 7 ^ (4 x 502) = 7 ^ 2008 = 01 (mod 100) Hence last 2 digits are 01