Q.$K_1$ and $K_2$ are the equilibrium constants for the reactions respectively.

[!FORMULA] $N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad (K_1)$

[!FORMULA] $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) \quad (K_2)$

What is the equilibrium constant for the reaction $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$

When several equilibria share species, their equilibrium constants combine algebraically in step with however the reactions themselves are added, reversed, multiplied or divided.

Reversing a reaction inverts its constant. If the forward reaction $xA+yB\rightleftharpoons lC+mD$ has equilibrium constant $K_C$, the reverse reaction $lC+mD\rightleftharpoons xA+yB$ has constant $K_C'=1/K_C$ (swapping numerator and denominator inverts the ratio). This is why the constant for the dissociation of one mole of $SO_3$ into $SO_2$ and $O_2$ is $(1/K_1)^{1/2}$, not simply $1/K_1$, when $K_1$ is defined for the two-mole formation reaction $2SO_2+O_2\rightleftharpoons2SO_3$: dissociating one mole of $SO_3$ is half of the reverse of the two-mole reaction, so its constant is the square root of the reverse constant, $\sqrt{1/K_1}$.

Adding reactions multiplies their constants. If $A\rightleftharpoons B$ has constant $K_1$ and $B\rightleftharpoons C$ has constant $K_2$, then the sum reaction $A\rightleftharpoons C$ (obtained by adding the two equations and cancelling the common intermediate B) has constant $K=K_1K_2$, because each intermediate species' concentration term cancels out of the product of the two separate expressions. Chained over three or more steps, this becomes $K_4=K_1K_2K_3$ for a four-species chain $A\rightleftharpoons B\rightleftharpoons C\rightleftharpoons D$ compared against the single direct step $A\rightleftharpoons D$. …