How does eve find the message m that Bob encrypted? The following is given:  


  $p=13$, $g=7$, $A=5$, $(c1,c2)=(10,8)$


I know $A=g^a \to a=3$  the message $m$ should be $c2*c1^{-a}$, which is $8\times 10^{-3} \mod 13$ but I don't know how to calculate the inverse part.
So what I get from your first computation is the following: you have looked for $a$ such that 
$A = g^a \mod 13$ and have found that $a = 3$. In your second step, you want to compute $$c_2\cdot c_1^{-a} = 8 \cdot 10^{-3} \mod 13.$$

Since $\text{gcd}(10,13) = 1$, we have that $10$ has a multiplicative inverse modulo $13$. This means that there is some $x$ such that 
$$10\cdot x \equiv 1 \mod 13$$
and we denote $x = 10^{-1}$. Therefore, your question becomes: 'find $x$ which satisfies this equation'. 

Okey, so we have the following equations:
$$\begin{align}
13 &amp;= 1 \cdot 10 + 3\\
10 &amp;= 3 \cdot 3 + 1\\
3 &amp;= 3 \cdot 1
\end{align}$$
this is Euclids division algorithm. From it , you can see that (starting from the second equation):
$$\begin{align}
1 &amp;= 10 - 3 \cdot 3\\
  &amp;= 10 - 3\cdot (13 - 1\cdot 10)\\
  &amp;= 4 \cdot 10 - 3 \cdot 13
\end{align}$$
and if we look at this modulo $13$, we see that 
$$1 = 4 \cdot 13 - 3 \cdot 13 \equiv 4 \cdot 10 \mod 13.$$
Therefore, we have shown that $10^{-1} \equiv 4 \mod 13$. Note that $10^{-3} \equiv 4^3 = 64 \equiv 12 \equiv -1 \mod 13$. 

Using all of this, we find that 
$$c_2 \cdot c_1^{-a} = 8\cdot 10^{-3} \equiv 8 \cdot (-1) \equiv -8 \equiv 5 \mod 13.$$

I hope this helps :)
To find $10^{-3}\mod 13$ we first note that it is equal to $\left(10^{-1}\right)^3\mod 13$.

We can find $10^{-1}\mod 13$ by the definition of inverses - we want to find an $x$ such that $x\times 10\equiv 1\mod 13$ where $x$ can also be denoted by $10^{-1}$

I used WolframAlpha to find  $10^{-1}=4\mod 13$, however you can do this by trying all values of $x$ from $2$ to $12$ until you find the desired value (there are far better methods for doing this btw, but beyond the scope of this answer). We note that $10$ will have an inverse modulo $13$ as $13$ is prime.

Therefore, \begin{align}10^{-3}\mod 13&amp;\equiv\left(10^{-1}\right)^3\mod 13\\
&amp;\equiv 4^3\mod 13\\
&amp;\equiv 64\mod 13\\
&amp;\equiv 12\end{align}