= Solution
First resolve the <orientation sign in quadrant harmonic spectral inversion>. The printed spectral definitions traverse the physical boundary clockwise, whereas both reconstruction rays run outward from zero. Integrating first in $k$ gives the kernel $i/(z-w)$. The resulting clockwise <Cauchy integral> is the negative of the desired <holomorphic function>. Thus, with exactly the spectral definitions used above, the consistent reconstruction is
$$
q_z(z)=-\frac1{2\pi}\left[\int_0^\infty e^{ikz}\widehat q_2(k)\,dk+
\int_0^{i\infty}e^{ikz}\widehat q_1(k)\,dk\right].
$$
The printed plus sign would reconstruct $-q_z$. For a concrete nonzero example, $q(z)=-2\operatorname{Re}[(1+i)/(z+1+i)]$ is real, <harmonic>, smooth on the closed quadrant and decays along both axes. With $\beta_1=\beta_2=0$, it has $h_1(0)=h_2(0)=1$ and $q_z=(1+i)/(z+1+i)^2$. The printed orientation returns the negative of this <derivative>, so the issue persists even under the stated corner condition.
Let $a=\tfrac12e^{-i\beta_1-i\gamma}$. In the expressions from the preceding part, the coefficient of $\Phi$ on the imaginary ray is $a$, and the coefficient on the real ray is $-a e^{4i\gamma}$. For each of the three allowed sums, $e^{4i\gamma}=1$. The unknown part is therefore proportional to
$$
\int_0^{i\infty}e^{ikz}\Phi(k)\,dk-\int_0^\infty e^{ikz}\Phi(k)\,dk=0.
$$
Indeed, close the first spectral quadrant. For $z=x+iy$ with $x,y>0$, $|e^{ikz}|=e^{-(x\operatorname{Im}k+y\operatorname{Re}k)}$, so the large-arc contribution vanishes for bounded $\Phi$. The <Cauchy integral theorem> proves the equality and eliminates the single unknown <function>.
Substituting only the known pieces leaves
$$
q_z(z)=\frac1{2\pi}\left[
e^{i\beta_2}\int_0^\infty e^{ikz}H_2(k)\,dk
-e^{i\beta_2+i\gamma}\int_0^\infty e^{ikz}H_1(-k)\,dk
-e^{-i\beta_1}\int_0^{i\infty}e^{ikz}H_1(k)\,dk
+e^{-i\beta_1-i\gamma}\int_0^{i\infty}e^{ikz}H_2(-k)\,dk
\right].
$$
This already determines the answer from $h_1,h_2$ alone. Evaluating the elementary ray kernels gives the more direct <rational-angle oblique derivative problem on a quadrant> formula
$$
\boxed{\begin{aligned}
q_z(z)=\frac{i}{2\pi}\bigg[
&e^{i\beta_2}\int_0^\infty\frac{h_2(x)}{z-x}\,dx
+e^{-i\beta_1-i\gamma}\int_0^\infty\frac{h_2(x)}{z+x}\,dx\\
&-e^{-i\beta_1}\int_0^\infty\frac{h_1(y)}{z-iy}\,dy
-e^{i\beta_2+i\gamma}\int_0^\infty\frac{h_1(y)}{z+iy}\,dy
\bigg].
\end{aligned}}
$$
No denominator vanishes for an interior point of the first quadrant.
For explicit formulas in the three cases, define
$$
I_2(z)=\int_0^\infty\frac{h_2(x)}{z^2-x^2}\,dx,\qquad
I_1(z)=\int_0^\infty\frac{h_1(y)}{z^2+y^2}\,dy,
$$
$$
K_2(z)=\int_0^\infty\frac{x h_2(x)}{z^2-x^2}\,dx,\qquad
K_1(z)=\int_0^\infty\frac{y h_1(y)}{z^2+y^2}\,dy.
$$
Combining the paired fractions and using $\beta_2=\gamma-\beta_1$ yields
$$
\boxed{q_z(z)=\frac{e^{-i\beta_1}}{\pi}
\begin{cases}
iz[I_2(z)-I_1(z)],&\gamma=0,\\
K_1(z)-K_2(z),&\gamma=\pi/2,\\
-iz[I_2(z)+I_1(z)],&\gamma=\pi.
\end{cases}}
$$
There is also a necessary <corner compatibility for collinear oblique derivative data>. When $\gamma=\pi/2$, the two prescribed <derivative> directions are opposites, so a continuous corner <gradient> requires $h_2(0)=-h_1(0)$. Together with the stipulated equality, this forces \b[$h_1(0)=h_2(0)=0$ in that case]. Existence was assumed, so admissible data must satisfy this additional consequence. Smoothness at the corner and decay at infinity also exclude the <homogeneous corner ambiguity in a quadrant Laplace problem>; no extra singular or growing term is permitted in the boxed reconstruction.
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