= Solution
Divide the two outer boundary equations respectively by $z$ and leave the derivative equation as written:
$$
A/z^2+B+Cz+Dz^3=0,\qquad
-A/z^2+B+2Cz+4Dz^3=0.
$$
Their difference gives $Cz+3Dz^3=2A/z^2$. As $z\to\infty$, the inner equations require $A,B\to1/4$. The two outer equations at leading order thus give $Cz+Dz^3=-1/4$ and $Cz+3Dz^3=0$, so
$$
C=-\frac3{8z}+O(z^{-2}),\qquad
D=\frac1{8z^3}+O(z^{-4}).
$$
The inner equations also give $A+B=1/2-C-D$ and $B-A=-2C-4D$. Solving and substituting yields
$$
\boxed{A=\frac14-\frac3{16z}+O(z^{-2}),\quad
B=\frac14+\frac9{16z}+O(z^{-2}),\quad
C\sim-\frac3{8z},\quad D\sim\frac1{8z^3}.}
$$
Replacing $z$ by $b/a$ gives all four requested approximations.
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