A couple of problems involving divisibility and congruence I'm trying to solve a few problems and can't seem to figure them out. Since they are somewhat related, maybe solving one of them will give me the missing link to solve the others.
$(1)\ \ $ Prove that there's no $a$ so that $ a^3  \equiv -3 \pmod{13}$
So I need to find $a$ so that $a^3 \equiv 10 \pmod{13}$. From this I get that $$a \equiv (13k+10)^{1/3} \pmod{13} $$ If I can prove that there's no k so that $ (13k+10)^{1/3} $ is a integer then the problem is solved, but I can't seem to find a way of doing this.
$(2)\ \ $ Prove that $a^7 \equiv a \pmod{7} $ 
If $a= 7q + r \rightarrow a^7 \equiv r^7 \pmod{7} $. I think that next step should be $ r^7 \equiv r \pmod{7} $, but I can't figure out why that would hold.
$(3)\ \ $ Prove that $ 7 | a^2 + b^2 \longleftrightarrow 7| a \quad \textbf{and} \quad 7 | b$
Left to right is easy but I have no idea how to do right to left since I know nothing about what 7 divides except from the stated. Any help here would be much appreciated. 
There're a lot of problems I can't seem to solve because I don't know how to prove that a number is or isn't a integer like in problem 1 and also quite a few that are similar to problem 3, but I can't seem to find a solution. Any would be much appreciated. 

For the first problem, an unimaginative but workable approach is just to check all the possibilities $a=0,1,2,\dots,12$. Do you see why, if these all fail, you're done? 
For the 3rd problem, I think you mean right implies left is easy. For left implies right, take each of the numbers $0,1,\dots,6$, square them, divide by 7, and note the remainders. From that you can work out all the possible remainders of $a^2+b^2$, and from that you can answer the question. 
Let me know if you've tried to work through this and found my hints too opaque. 
