How does law of quadratic reciprocity work? From the book,
Suppose $p \equiv 1 \pmod{4}$, then by law of quadratic reciprocity, we have:
$$\left(\frac{3}{p}\right)  = \left(\frac{p}{3}\right) $$
Next, if $p \equiv 2 \pmod{3}$, then $p \equiv 5 \pmod{12}$
Hence, $$\left(\frac{3}{p}\right)  = \left(\frac{p}{3}\right) = -1$$
How do they get those Legendre fraction equal to $-1$?
From my understanding:
$$\left(\frac{q}{p}\right)  \cdot \left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$$
So $q = 3 \implies \frac{q-1}{2} = \frac{3-1}{2} = 1$
For $p$, I take $p \equiv 5 \pmod{12} \implies p = 5 + 12k$, for some integers k.
Hence, $\frac{p-1}{2} = \frac{12k + 5 - 1}{2} = \frac{12k + 4}{2} = 6k + 2.$
And this $6k + 2$ is even :( ! How does $(-1)^{even} = -1$?
Any idea? I think I made some logic mistakes somewhere, but I couldn't find where.
Update
The problem was from Elementary Number Theory and Its Application - Kenneth H.Rosen 5th Edition.  

Problem
  Using the law of quadratic reciprocity, show that if $p$ is an odd prime, then
  $$\left(\frac{3}{p}\right) = 1 \text{  if  } p \equiv \pm 1 \pmod{12}$$
  $$\left(\frac{3}{p}\right) = -1 \text{  if  } p \equiv \pm 5 \pmod{12}$$
Solution
  
  Thanks, 

Now I'm even more confused :(!
Consider two cases:
Case 1
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 1 \pmod{3}$$
Then,
$$p \equiv 1 \pmod{12} \implies p = 12k + 1$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k}{2} = 6k = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$. 
Case 2
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 2 \pmod{3}$$
Then,
$$p \equiv 5 \pmod{12} \implies p = 12k + 5$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k + 4}{2} = 6k + 2 = 2(3k + 1) = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$. 
I don't see how these arguments can be deduced to the solution. Any suggestion?   

If $p\equiv 2 \mod 3$ then $p$ is not a square mod $3$. That is the justification for the second line. 
As you correctly calculated, if $p$ is $5 \mod 12$ (and $q$ is odd) then $(-1)^{\frac {p-1}{2} \cdot \frac {q-1}{2}}$ is $(-1)^{\text{even}} = 1$.  
You have an incorrect statement of the main case of quadratic reciprocity, for $p$ and $q$ odd primes. Instead of an equal sign between $(\frac p q)$ and $(\frac q p)$ there should be no symbol, or perhaps $\times$. The product $(\frac p q) (\frac q p)$ equals $(-1)^{\frac {p-1}{2} \cdot \frac {q-1}{2}}$, which is to say the product is $1$ except when $p$ and $q$ are both $3 \mod 4$, in which case it is $-1$. 
