There are two methods 

Method 1: try to find some pattern

7¹= 07
7²= 49
7³= 343
7⁴= 2401
7⁵= 16807
7⁶= 117649
7⁷= 823543
7⁸= 5764801

We can see that last 2 digits are repeating every 4 times.
so we can say that 
last two digit in 7ⁿ=07 when n=4k+3
last two digit in 7ⁿ=49 when n=4k+2
last two digit in 7ⁿ=43 when n=4k+1
last two digit in 7ⁿ=01 when n=4k

Hence last two digit in 7²⁰⁰⁸ is 01.

Method 2:

For finding last two digit of odd number(in our case 7 is odd). 
we will reduce the base to a number with unit digit 1.

7²⁰⁰⁸ = 49¹⁰⁰⁴ = 2401⁵⁰²

Now for base ending with 1.
formula is ([digit at tens place of base multiplied by last/unit digit of power] * 10 + 1)
In our case tens place of base is 0 and unit digit of power is 2
therefore it will be ([0*2]*10 + 1) = (0*10 + 1) = 1
Hence last two digit in 7²⁰⁰⁸ is 01.