Rate expression for the final product
I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that 

$$k_2[B]^2[C]=k_3[D][E]$$

This approximation is why the teacher is always saying that only the slow reaction is important 

Fast equilibrium means that the equilibrium is quickly governed by thermodynamics laws and so the constant

$$K = \frac{[B]}{[A]}$$

is not variating, so you can express $B$ in function of $A$

$$\frac{d[F]}{dt} = k_2K^2[A]^2[C]$$

I don't know how far you need to go but if you really want to solve for $[F]$ you need to write an equation of conservation

$$[N] = [A] + [B] + [C] + [D] + [E] + [F] = cst$$
$$d[A] + d[B] + d[C] + d[D] + d[E] + d[F] = 0$$

$$d[D] = 0$$
$$d[E] = 0$$

You can also say that $d[C] \approx 0$ because it looks like it's just a partner of collision, (it's the solvent) because the solvent is in much larger quantity you can say the concentration stay pretty the same

$$d[A] + d[B] = - d[F]$$

$$[A] - [A]_0 + [B] = -[F]$$

$$[A](1+K) = [A]_0 - [F]$$

$$[A]^2= \frac{([A]_0 - [F])^2}{(1+K)^2}$$

$$\frac{d[F]}{dt} = K_0\frac{([A]_0-[F])^2}{(1+K)^2}$$

with $K_0 = k_2K^2[C]_0$