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:
$$\ce{CH2(CN)2 ->[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$
$\ce{CH(CN)2-}$ can be consumed by two pathways:
$$\ce{CH(CN)2- + Br2 ->[k_2] BrCH(CN)2 + Br-} \\ \ce{CH(CN)2- + H+ ->[k_{-1}] CH2(CN)2} \\ \mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$
Thus:
$$\dfrac{d[\ce{B}]}{dt}=k_1[\ce{A}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$
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 $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&\ [\ce{H+}]$, 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.