Show that this mechanism is consistent with the rate law - how to?
You have a three step mechanism where only one step is slow and the other steps are fast.

Since the first two steps (which are, in fact equilibria) are fast, it can be assumed that each of the equilibria holds throughout the reaction. Therefore, the expression of the equilibrium constant for each step can be used:
$$\ce{K_1=\frac{[I]^2}{[I2]} \,\,\, and \,\,\, K_2=\frac{[H2I]}{[H2][I]}}$$
So, $\ce{[I]=\sqrt{\ce{K_1[I_2]}}\,\,\,\,and,\,\,[H2I]=K_2[H2][I]=K2[H2]\sqrt{\ce{K1[I2]}}}$

As only the last step is slow, the rate of formation of the product (which is the rate of the overall reaction) only depends on the last step.

$$\ce{rate=\frac{1}{2}\frac{d[HI]}{dt}=k_s[H2I][I]}$$
Here, $\ce{k_s}$ is the rate constant of the last, slow step. Substituting the expressions for $\ce{[H2I]}$ and $\ce{[I]}$ in the rate law should give you an expression of the form $\ce{rate=constant\times[H2][I2]}$, which is what the given rate law is.