What is $5^{-1}$ in $\mathbb Z_{11}$? I am trying to understand what this question is asking and how to solve it. I spent some time looking around the net and it seems like there are many different ways to solve this, but I'm still left confused.

What is the multiplicative inverse of $5$ in $\mathbb Z_{11}$.
      Perform a trial and error search using a calculator to obtain your answer.

I found an example here:

In $\mathbb Z_{11}$, the multiplicative inverse of $7$ is $8$ since $7 * 8 \equiv 56 \pmod {11}$.

This example is confusing to me because $1 \pmod {11} \equiv 1$. I don't see how $56$ is congruent to $1 \pmod {11}$. 

You may want to read up on modular arithmetic in order to approach this problem. For every numbers $k$ and $n$, there are infinitely many numbers $m$ such that $m\equiv k (\mod n)$, because $m=k+nt$ works for every $t$!
In $\mathbb{Z}_{11}$ (note: the notation $\mathbb{Z}/11\mathbb{Z}$ is generally considered slightly better), our elements are actually equivalence classes:
$[0]\in\mathbb{Z}_{11}$ is the set of all numbers $m$ such that $m=11t$ for some $t\in\mathbb{Z}$, $[1]\in\mathbb{Z}_{11}$ is the set of all numbers $m$ such that $m=11t+1$ for some $t\in\mathbb{Z}$, and so on.
Your question is asking you to look at the representatives $x\in\{0,1,2,3,4,5,6,7,8,9,10\}$ and try to choose $x$ so that $5x=11t+1$ for some $t\in\mathbb{Z}$; i.e. find $x$ so that $5x\equiv 1(\mod 11)$. In particular, it wants you to just multiply each $5$ by each number in the above set, and then figure out which of these results can be written in the form $11t+1$ for some $t\in\mathbb{Z}$.
