Deriving fluorescence intensity equations
Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then 

$$\begin{align}  &amp;I_O \quad \rightleftharpoons  &amp;I_D \\ &amp;1-\alpha &amp;\alpha \end{align}$$

and the equilibrium constant $K=I_D/I_O=\alpha/(1-\alpha)$.

The total emission is $\displaystyle F=I_O(1-\alpha)+I_D\alpha=I_O\frac{1}{1+K}+I_D\frac{K}{1+K}$

Rearranging this gives $\displaystyle \frac{1}{I_O+KI_D}+\frac{1}{I_D}=\frac{1}{F}$

Replacing intensity with product of fluorescence yield and concentration and using the rather odd definitions given  should give the equation you seek. Note that $F-\phi_{AO}C_{AO}$ is the intensity of the bound dye.

The first question states that the intensity due to the bound dye $F-\phi_{AO}C_{AO}$ is equal to the amount bound times the difference in yields, this because the dye fluoresces at the same wavelength whether bound or not.