How can I solve these congruences? 
I have no idea, how to solve these congruences if you can help me please. Thanks a lot.

Notice $2$ is a primitive root modulo $11$.  In other words, any number between $1$ and $10$ is congruent modulo $11$  to a unique power $2^k$ for $1 \leq k \leq 10$.  Notice $10 \equiv -1$, so on your first congruence you're really trying to solve $x^5 \equiv -1 \pmod {11}$.  Let's suppose $2^k$ is a solution for $x$ (so to find $x$, you'll have to solve for $k$), so $2^{5k} \equiv -1 \pmod{11}$.  Also $-1 \equiv 2^5$, so then $2^{5k} \equiv 2^{5}$.  But this is the same as saying that $$5k \equiv 5 \pmod{10}$$ Do you see why?  Also, note that now we have to work modulo $\varphi(11) = 10$ rather than mod $11$.  Anyway, it's easy to see that $k$ solves that congruence if and only if $k$ is odd.  So $k = 1, 3, 5, 7, 9$ are all the solutions to the congruence $5k \equiv 5 \pmod {10}$, so $$x = 2, 2^3, 2^5, 2^7, 2^9$$ will be all the solutions to the congruence $10x^5 \equiv 1 \pmod {11}$.  
For the other problem, find a number $g$ such that $5g \equiv 1 \pmod {11}$ (for example, $g = 9$).  Then the congruence $5x^5 \equiv 3 \pmod{11}$ is equivalent to the congruence $x^5 \equiv 27$, or $x^5 \equiv 5 $.  Apply the same type of argument as the first question.  Ask me if there's anything you don't understand about what I said.  
