= Solution
Split the <Fourier transform> integral at zero:
$$
\boxed{\widehat g(k)=\int_0^\infty e^{-(\lambda+ik)x}dx+\int_0^\infty e^{-(\lambda-ik)x}dx=\frac{2\lambda}{\lambda^2+k^2}.}
$$
By <Fourier inversion> applied to this pair, or by the <residue theorem>, the other transform is
$$
\boxed{\widehat h(k)=\frac\pi\mu e^{-\mu|k|}.}
$$
For example, inversion with $\lambda=\mu$ gives $\int e^{ikx}(k^2+\mu^2)^{-1}dk=(\pi/\mu)e^{-\mu|x|}$; evenness supplies the desired sign convention.
Take the spatial <Fourier transform> of the <Laplace equation>. For $y>0$ it gives $\widehat\psi_{yy}-k^2\widehat\psi=0$. Removing the exponentially growing solution and matching the boundary gives
$$
\widehat\psi(k,y)=\widehat f(k)e^{-|k|y}.
$$
At $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifies
$$
\mathcal F^{-1}[e^{-|k|y}](x)=\frac1\pi\frac{y}{x^2+y^2}=P_y(x),
$$
the <Poisson kernel for the upper half-plane>. The <convolution theorem> therefore gives the <Poisson integral>
$$
\boxed{\psi(x,y)=\frac y\pi\int_{\mathbb R}\frac{f(u)}{(x-u)^2+y^2}du.}
$$
This is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero <harmonic function> $e^{x^2-y^2}\sin(2xy)=\operatorname{Im}e^{(x+iy)^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to\infty$, but grows rapidly in $x$ and has no ordinary spatial <Fourier transform>. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.
For the interval data, direct integration gives the <Poisson extension of interval boundary data>
$$
\boxed{\psi(x,y)=\frac c\pi\left[\arctan\frac{a-x}{y}+\arctan\frac{a+x}{y}\right],\qquad y>0.}
$$
As $y\downarrow0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi\sim2ac/(\pi y)$ for fixed $x$ and large $y$.
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