For part (a), similar to problem 6, we get $$\hat{u}(k,y) = A(k)e^{-|k|y}+B(k)e^{|k|y}$$
Because we have two boundary conditions $$\hat{u|}_{y=0} = \hat{f}(k)\\\hat{u|}_{y=1} = \hat{g}(k)$$
We cannot discard anything. Then,$$A(k) + B(k) = \hat{f}(k)\\A(k)e^{-|k|}+B(k)e^{|k|} = \hat{g}(k)$$
Then we get, $$A(k) = \frac{e^{|k|}\hat{f}(k)}{2\sinh(|k|)} - \frac{\hat{g}(k)}{2\sinh(|k|)}\\B(k)=-\frac{e^{-|k|}\hat{f}(k)}{2\sinh(|k|)}+\frac{\hat{g}(k)}{2\sinh(|k|)}$$
Hence, we get$$\hat{u}(k,y) = \frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}\\u(x,y) = \int_{-\infty}^{\infty} [\frac{\sinh(|k|(1-y))\hat{f}(k)}{\sinh(|k|)}+\frac{\sinh(|k|y)\hat{g}(k)}{\sinh(|k|)}]e^{ikx} dk$$
