Two different methods to express the equilibrium constant give different results
So which one is actually right and which one is wrong? Why?


The second method is wrong.  In your second method, you've written: $\ce{I2 \+ Br2 -&gt;IBr  + IBr }$ and then assigned the concentration $x$ and $x$ to each of the two (identical!) $\ce{IBr}$'s. Till now, it's fine. 

The $K_\mathrm c$ has been used by you as:

$$K_\mathrm c=\frac{[\ce{IBr}][\ce{IBr}]}{[\ce{I2}][\ce{Br2}]}$$

which is okay but a better way to write it would have been 

$$K_\mathrm c=\frac{[\ce{IBr}]^2}{[\ce{I2}][\ce{Br2}]}$$

as you've taken the single concentration term for $\ce{IBr}$ (=$2x$) from the correct formula, split it up into two parts ($x$ and $x$), and then used those two parts separately in $K_\mathrm c$, as if there were not one but two distinct $\ce{IBr}$'s being formed! It is just confusing.

Anyways, the first formula will still work. The main problem here is that $[\ce{IBr}]=\color{blue}{\text{total }}\text{concentration of}~\ce{IBr}$. When you put any reactant/product inside the square brackets, you're supposed to write their total concentration, even if they have been split up in parts. So, when in the derivation of $K_\mathrm c$, you write:

$$K_\mathrm c = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^c.[\ce{D}]^d}{[\ce{A}]^a.[\ce{B}]^b}$$

for the reaction: $$\ce{aA + bB —&gt; cC + dD}$$

each of those individual $[\ce{C}]$'s will refer to the total concentration of $\ce{C}$. That means, as $\ce{C}$ was separately written $c$ times by you in the reaction, the final $[\ce{C}]$ that you use will be the sum of all those $c$ $[\ce{C}]$s. 

So, in the end of your second method, while applying the correct formula for $K_\mathrm c$, you must add those $x$ and $x$ together, and then write:

$$K_\mathrm c=\frac{(x+x)\times(x+x)}{(2-x)\times(2-x)}$$

into the formula you were using.