The digit in the unit's place of the number represented by $$(7^{95}-7^{58})$$ is

$$(7^{95}-7^{58})$$ = N

The unit digit of {$$7^1,7^2,7^3,7^4$$} = {7, 9, 3, 1} . And this cycle gets repeated for every 4 powers ..

So if the exponent is in the form of :
1. form of {4k} : unit digit is 1
2. form of {4k + 1} : unit digit is 7
3. form of {4k + 2} : unit digit is 9
4. form of {4k + 3} : unit digit is 3

Now, 95 = 4k + 3 , so the unit digit would be 3 .
         58 = 4k + 2 , so the unit digit would be 9.

Therefore, for this $$(7^{95}-7^{58})$$ = {Number with unit digit 3 - Number with unit Digit 9} 
                                                           = {Number with unit digit of (13-9)}
                                                           = 4.