We also know these two measures are finite.
The hint to this question says to use the Radon-Nikodym theorem, but it seems that either they are referencing an alternate phrasing that I cannot find, or I am entirely unsure how this helps us. Firstly, I know that to use R-N I need to have one measure be absolutely continuous with respect to the other. So, with that in mind, I want to come up with some new measure defined from the others. I am unsure how to concoct such a measure in any meaningful way here.
Write $\mu=\lambda+\tau$ with $\lambda << \nu, \tau\perp \nu$. Let $g=\frac {d\lambda} {d\nu}$. Let $B$ be measurable set such that $\nu (B)=0$ and $\tau (B^{c})=0$. Let $f=\frac g {1+g} 1_{B^{c}}+1_B$. Note that $$\int_A(1-f)d\mu$$ $$=\int_A(1-f)d\lambda+\int_A(1-f)d\tau$$ $$=\int_{A}(1-f)d\lambda+\int_{A\cap B}(1-f)d\tau$$ and the second term is $0$. Finally, (since $\lambda (B)=0$ we get) $$\int_A(1-f)d\mu$$ $$=\int_{A}(1-f)d\lambda=\int_{A\cap B^{c}}(1-f)gd\nu$$ $$=\int_{A\cap B^{c}} fd\nu(\equiv \int_{A} fd\nu)$$ since $(1-f)g=f$ on $B^{c}$.