How does changing the concentrations of intermediate products/reactants change the rate law?
The steady state approximation allows us to assume that the concentration of $\ce{B}$ does not change.

$$\dfrac{d[\ce{B}]}{dt}=0$$

Thus the rates of production and consumption of $\ce{CH(CN)2-}$ must be equal.


  $\ce{CH(CN)2-}$ can be produced by one pathway:


$$\ce{CH2(CN)2 -&gt;[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$  


  $\ce{CH(CN)2-}$ can be consumed by two pathways:


$$\ce{CH(CN)2- + Br2 -&gt;[k_2] BrCH(CN)2 + Br-} \\ 
\ce{CH(CN)2- + H+ -&gt;[k_{-1}] CH2(CN)2} \\
\mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$


  Thus:


$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$

Now you can can solve for $[\ce{CH(CN)2-}]$ and substitute into the rate law equation for the second step:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=k_2[\ce{CH(CN)2-}][\ce{Br2}]$$

You should get a final rate law that contains $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&amp;\ [\ce{H+}]$, and then you can answer your questions.


  Update based on revised question.


Your new rate late is correct:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=\dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]+k_{-1}[\ce{H+}]}$$

Your interpretation of the simplification you can do is correct for one case and almost correct for the other.


  $[\ce{H+}] \ll [\ce{Br}]$


In this case, we assume that $k_{-1}[\ce{H+}]\ll k_2[\ce{Br}]$ so 

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_2[\ce{Br2}] \\
\therefore \ 
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]}\approx k_1[\ce{CH2(CN)2}$$

which is as you figured.


  
    $[\ce{H+}] \gg [\ce{Br}]$
  


In this case, we assume that $k_{-1}[\ce{H+}]\gg k_2[\ce{Br}]$ so 

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_{-1}[\ce{H+}] \\
\therefore \ 
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_{-1}[\ce{H+}]}$$

Note that this approximation does not remove the $k_2[\ce{Br2}]$ term in the numerator.