The temperature $T(x,y)$ of a long rectangular bar of X-sectional width $a$ and depth $b$ satisfies the equation $\nabla ^2 T=0$, and subject to the boundary conditions
$$T(0,y)=T(a,y)=0,$$
$$T(x,0)=0, \;\;T(x,b)=T_b(x)$$
Use the method of separation of variables to express the solution to this problem in the form
$$T(x,y)=\int^a_0 G(x,/xi;y)T_b(\xi)d\xi,$$
where the function $G$ is to be determined.
Assume $T(x,y)=X(x).Y(y) $
$$\nabla^2 T=0 \rightarrow \frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2}=0$$
$$\frac{X''}{X}=\frac{Y''}{Y}=-m^2$$
\begin{align} X&=A\cos{mx}+B\sin{mx},\\ Y&=C \cosh{mt}+D\sinh{mt} \end{align} \begin{align} T(0,y)&=0 \rightarrow X=B\sin{mx}\\ T(a,y)&=0 \rightarrow m=\frac{n\pi}{a} \\ T(x,0)&=0 \rightarrow Y=D \sinh{mt} \end{align}
$$ \therefore T_b(x)=B\sin{\frac{m\pi x}{a}}\cdot D\sinh{\frac{m\pi t}{a}} $$
$$T(x,y)=\sum ^{\infty}_{n=1} E_n T_b(x)$$
Then how do I determine G?
There are a few things to notice. Taking many shortcuts in the process doesn't usually pay; I know it doesn't for me. Starting with $T(x,y)=X(x)Y(y)$, $$ \nabla^2T=X''(x)Y(y)+X(x)Y''(y)=0 \\ \frac{X''Y}{XY}+\frac{XY''}{XY}=0 \\ \frac{X''}{X}+\frac{Y''}{Y}=0 \\ \frac{X''}{X}=-\frac{Y''}{Y} \\ \frac{X''}{X} = -\lambda,\;\;\; \lambda = \frac{Y''}{Y}. $$ Skipping steps led you to conclude that the equations in $X$ and $Y$ are identical. They are not. The conditions $$ T(0,y)=T(a,y)=0 \implies X(0)=X(a)=0. $$ That can happen only for trigonometric functions, not the hyperbolic functions, which means $\lambda > 0$, and $X(0)=0$ gives $$ X_{\lambda}(x)=A\sin(\sqrt{\lambda}x) $$ And $X(a)=0$ means $\sin(\sqrt{\lambda}a)=0$ so that $$ \sqrt{\lambda}a = n\pi,\;\;\; n=1,2,3,\cdots \\ \lambda = \frac{n^2\pi^2}{a^2},\;\; X_n(x) = A_n\sin(n\pi x/a). $$ Then $$ Y_{n}(y) = B_n\sinh(n\pi y/a)+C_n\cosh(n\pi y/a) $$ However, $Y_n(0)=0$ gives $C_n=0$, which leads to a solution $$ T(x,y)=\sum_{n=1}^{\infty}D_n\sin(n\pi x/a)\sinh(n\pi y/a) $$ The remaining condition at $y=b$ determines the constants $D_n$: $$ T_b(x) = T(x,b) = \sum_{n=1}^{\infty}D_n\sin(n\pi x/a)\sinh(n\pi b/a). $$ Multiplying by $\sin(n\pi x/a)$, integrating both sides over $[0,a]$, and using the orthogonality of the sin functions on $[0,a]$ results in equations for the $D_n$: \begin{align} \int_{0}^{a}T_b(x)\sin(n\pi x/a)dx&=D_n\sinh(n\pi b/a)\int_{0}^{a}\sin^2(n\pi x/a)dx \\ &=D_n\sinh(n\pi b/a)\frac{1}{2} \end{align} Therefore, $$ T(x,y)=\sum_{n=1}^{\infty}\left[\frac{2}{\sinh(n\pi b/a)}\int_{0}^{a}T_b(x')\sin(n\pi x'/a)dx'\right]\sin(n\pi x/a)\sinh(n\pi y/a) \\ = \int_{0}^{a}\left[\sum_{n=1}^{\infty}\sin(n\pi x'/a)\sin(n\pi x/a) \frac{2\sinh(n\pi y/a)}{\sinh(n\pi b/a)}\right]T_b(x')dx' $$ The kernel is enclosed by square brackets in the final expression.