1. The proposed three-step mechanism for 2NO + 2H2 ---> N2 + 2H2O is as follows:
2NO <---->  N2O2  (step 1; fast equilibrium)
N2O2 + H2  ---->  N2O + H2O  (step 2; slow)
N2O + H2  ---->  N2 + H2O (step 3; fast)
Since the second step is rate-determining, the overall reaction rate is governed by the rate of this step. Therefore,
-(1/2)d[NO]/dt = k2[N2O2][H2]   ....
Note that the first step is a fast equilibrium process, and
k1[NO]^2 = k-1 [N2O2]  ....
where k1 and k-1 are the forward and backward rate constants. Substituting  into  we obtain
-(1/2) d[NO]/dt = k2*(k1/k-1)[NO]^2 [H2]
-d[NO]/dt = 2k2*(k1/k-1) [NO]^2[H2] = k[NO]^2[H2]
where k = 2k1k2/k-1 is the effective rate constant.

2. This question can be solved using the same technique.
The proposed mechanism is
I2  <---->  2I  (step 1; fast equilibrium)
2I + H2  ---->  2HI  (step 2; slow)
Since the second step is rate-determining, then
-d[I2]/dt = k2[I]^2 [H2]   ....
From step 1, we know that
k1[I2] = k-1[I]^2  ....
where k1 and k-1 are the forward and backward rate constants. Putting  into  we get
-d[I2]/dt = k2*(k1/k-1) [I2][H2] = k[I2][H2]  ....
where k = k1k2/k-1 is the effective rate constant.
We can easily see that rate expression  is first-order w.r.t. [I2] and [H2], respectively, as required.

I will propose the following mechanism:
O3  <---->  O2 + O (step 1; fast equilibrium)
O3 + O  ---->  2O2 (step 2; slow)
The rate of the reaction can be described as:
rate = k2[O3][O]  ....
since the second step is rate-determining. Now, using the first equilibrium, we can write
k1[O3] = k-1[O2][O]
[O] = k1[O3]/k-1[O2]  ....
Substituting  into , we will get the rate expression
rate = k2[O3](k1[O3]/k-1[O2]) = k1k2[O3]^2/k-1[O2]
or
rate = k[O3]^2/[O2]
where k = k1k2/k-1.

Whenever you see the inverse dependence of the rate on certain species, it is likely that the species is involved in the reverse process of the pre-equilibrium step.                                                                                                                                                                                                                                                                                                                       

Which question? 6, 7 or 8?

OK, let me try  of question 8.

Remember that the rate expression is always determined, in approximation, by the slow reaction step, i.e., the rate-limiting step.

A. A single-step reaction, and thus the rate expression is
rate = k[CO][NO2]

B. The given reaction step is a rapid equilibrium. Using only this we cannot formulate the rate expression. Hence I assume that the following slow reaction is as such:
N2O4 + 2CO  ----> 2CO2 +2NO
The rate expression is thus rate = k2[N2O4][CO]^2
Note that k1[NO2]^2 = k-1[N2O4]
So, rate = (k1k2/k-1)[NO]^2[CO]^2

C. Since the slow reaction is 2NO2  ---->  N2 + 2O2, the rate expression is rate = k[NO2]^2

D. Since there are two slow steps in the mechanism, the overall rate of reaction is governed by both steps, and the rate expression is thus
rate = k1[N2O4][CO]^2 + k2[NO2]^2

So, you can see that only the mechanism C yields the rate expression in the required form of rate expression.