Let's attack 4^34 first:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256 ---> This indicates that if the exponent is odd, the last digit will be a 4; if the exponent is even, the last digit will be a 6.
Now 7^35:
7^1 = 7 7^5 = 16807
7^2 = 49 7^6 = 117649
7^3 = 343 7^7 = 823543
7^4 = 2401 7^8 = 5764801
This indicates that it cycles in groups of 4, so you can keep doing this until you get to 7^35. (You really don't have to keep doing the multiplications.)
Another way is to divide the exponent by 4:
if the remainder is 1, the answer ends in a 7;
if the remainder is 2, the answer ends in a 9;
if the remainder is 3, the answer ends in a 3; and
if the remainder is 0, the answer ends in a 1.
Dividing 35 by 4, the remainder is 3; therefore the last digit is a 3.
Now, the problem is reduced to adding 1 to a number whose last digit is 6 to another number whose last digit is 3. So what is the last digit of 1 + 6 + 3?