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. 

HINT $\rm\ (2)\quad\ mod\ 7\!:\ \{\pm 1,\:\pm 2,\:\pm3\}^3\equiv\: \pm1\:,\:$ so squaring yields $\rm\ a^6\equiv 1\ \ if\ \ a\not\equiv 0\:.$  
$\rm(3)\quad \ mod\ 7\!:\ \ if\ \ a^2\equiv -b^2\:,\:$ then, by above, cubing yields $\rm\: 1\equiv -1\ $ for $\rm\ a,b\not\equiv 0\:.$ 
$\rm(1)\quad \ mod\ 13\!:\ \{\pm1,\:\pm3,\:\pm4\}^3 \equiv \pm 1,\ \ \{\pm2,\pm5,\pm6\}^3\equiv \pm 5\:,\: $ and neither is $\rm\:\equiv -3\:.$
If you know Fermat's little theorem or a little group theory then you may employ such to provide more elegant general proofs - using the above special cases as hints.
