As usual, let's setup our partial fourier transform, shifting from x basis to k basis, i.e.
$$ \hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx} f(x) dx $$
Thus equations (1,2,3) become
$$ \hat{u_{yy}} - k^2 \hat{u} = 0 , \hat{u(k,0)} = 0 , (\hat{u_y}(k,1) + \alpha \hat{u}(k,1)) = \hat{g}(k)$$
Now, the fourier transform g(x) into $\hat{g}(k) $ is simply:
$\hat{g}(k) = \frac{1}{2\pi} \int_{-\infty}^{infty} e^{-ikx} g(x) dx = \frac{1}{2\pi} \int_{-1}^{1} e^{-ikx} dx = \frac{1}{2\pi} \frac{-e^{-ikx}}{ik} |_{-1}^{1} = \frac{1}{2\pi} \frac{e^{ik} - e^{-ik}}{ik} = \frac{1}{\pi k} sin(k)$
The solution to the differential equation from the fourier transform of (1) is:
$\hat{u} = A(k)e^{|k|y} +B(k)e^{-|k|y} $ , applying (2), we need: $-B(k) = A(k).$
$\hat{u} = A(k)e^{|k|y} -A(k)e^{-|k|y} = A(k)\sinh(|k|y)$
Finally, applying the condition (3):
$(A(k)|k| \cosh(|k|) + \alpha A(k)\\sinh(|k|) = \frac{sin(k)}{\pi k}$
So that: $A(k) = \frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1} $
Thus, our solution in x basis is given as:
$f(x) = \int_{-\infty}^{\infty} \hat{f(k)} e^{ikx} dk = \int_{-\infty}^{\infty} (\frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1})\sinh(|k|y)e^{ikx}dk $
