a=1+4 at power 34+7 at power 35.What's the last digit of a?

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? 

$$\\1+4^{34}+7^{35}\\
4^n=4,16,64,256,.... $ when n is even it ends in 6$\\
7^n=7,49,343,2401, 16807, $ So the last digits are 7,9,3,1,7,...$\\
35=4*8+3$ so the last digit of $ 7^{35}$ is 3 (3 is the third number in the sequence)$\\
$Edited: Geno is right, these three digits need to be added$\\\\
$ The last digit will be hte last digit of $ 6+3+1 $ Which is zero $$$

Fixed Gino is correct. (So am I now)