using the order of operations, how do i get 5 as an answer using the number 8 only 4 times?

the number 8 can only and must be used 4 times. i'm trying to get an answer of 2 for one problem and 5 for the other
 

-2*8 + 81/3 - 8 + (8-4)2 + 11
 
first:  parentheses.. (8-4) = 4
 
= -2*8 + 81/3 -8 + 42 + 11
 
2nd:  exponents:  81/3 = 2, 42=16
 
= -2*8 +2 - 8 + 16 + 11
 
3rd: multiplication / division left to right
 
= -16 + 2 - 8 + 16 + 11
 
4th:  addition / subtraction left to right
 
= 5
 
you can try to set one up to get a total of 2 as the final answer
 

hi, suzy!
 
i love these kinds of problems!  they always remind me of the math game 24 where you are given 4 numbers and the objective is to use those 4 numbers (only once) and addition, subtraction, multiplication, and division to equal 24.  great game (at least for a math nerd)!
 
but back to the topic at hand.
 
to get the answer of two, you can try working backwards.  brainstorm some ways to get two normally (ie - 2*1 or 1+1)  getting a 2 with two 8s would be difficult, but you can get 1 with two 8s by dividing them.  so,
1 + 1 becomes (8 ÷ 8) + (8 ÷ 8) = 2
 
regarding the answer 5, i'm not sure you can get that answer by starting from four 8s.  you can either have a starting pair of 8s that give you 0 (subtraction), 1 (division), 16 (addition), or 64 (multiplication).  0 doesn't get you anywhere - it just leaves you with trying to get 5 with two 8s instead of 4.  16, and 64 are both even numbers like 8, so if you add, subtract, or multiply by 8 you will get an even number.  in this case, if you divide 16 or 64 by 8, you also end up with an even number.  and when you start out by dividing two 8s to give you 1, i can't find a permutation that can give you 5 from a 1, 8, and 8.