Find the Green's function $G(\mathbf{x},\xi)$, such that $\nabla^2G = \delta(\mathbf{x}-\xi)$

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a) Consider a domain

$$D=[(x.y)|x\gt0,y\gt 0]$$

Let $\mathbf{x}=(x,y)$ and $\xi = (\xi_{x},\xi_{y})$.

Find the Green's function $G(\mathbf{x},\xi)$, such that $$\nabla^2G = \delta(\mathbf{x}-\xi), \mathbf{x}\in D$$

subject to

$$\frac{\partial G}{\partial x}(0,y,\xi)=0\space\space\space \text{for} \space\space\space y\gt0$$

and

$$G(x,0,\xi)=0\space\space\space \text{for} \space\space\space x\gt 0$$

b) Using part a), solve $$\nabla^2{u}=0, \textbf{x} \in D$$

subject to $u$ tends to $0$ as $|\textbf{x}|$ tends to $\infty$

$$\frac{\partial u}{\partial x}(0,y)=h(y)\space\space\space \text{for} \space\space\space y\gt0$$

and

$$u(x,0)=g(x)\space\space\space \text{for} \space\space\space x\gt0$$

So i'm pretty sure we use the method of images for a). Let the source be $\xi=(\xi_{x},\xi_{y})$, then the image sources are $\xi_{1}=(-\xi_{x},\xi_{y})$,$\xi_{2}=(\xi_{x},-\xi_{y})$,$\xi_{3}=(-\xi_{x},-\xi_{y})$.

So we have $\nabla ^2 G=\delta ( \textbf x - \xi )+ \delta ( \textbf x - \xi_1 ) - \delta ( \textbf x - \xi_2 )- \delta ( \textbf x - \xi_3 )$

Therefore $G(\textbf x , \xi)=\frac1{2 \pi} \ln |\textbf x - \xi |-\frac1{2 \pi} \ln |\textbf x - \xi_1 |-\frac1{2 \pi} \ln |\textbf x - \xi_2 |+\frac1{2 \pi} \ln |\textbf x - \xi_3 |$

I am completely lost after this. How would I go about using this for solving b)?