Three. (25 points) Given the system of equations about $x$ and $y$:
$$
\left\{\begin{array}{l}
k x^{2} + y + (k - a) = 0, \\
y = -(k + a) x + b c
\end{array}\right.
$$

It has only one integer solution, where $k, a, b, c$ are all integers, and $a > 0, a, b, c$ satisfy $a^{2} - a + b c = -1, b + c = 2$. Find:
(1) The value of $a$;
(2) The value of $k$ and the corresponding solution $(x, y)$.
(1) From $b+c=2, bc=a^{2}-a+1$, it is easy to know that $b, c$ are the roots of the equation $x^{2}-2x+a^{2}-a+1=0$.
Since $b, c$ are integers, we have
$\Delta=(-2)^{2}-4\left(a^{2}-a+1\right) \geqslant 0$.
Thus, $a^{2}-a \leqslant 0$.
Also, $a>0$, so $a \leqslant 1$.
Since $a$ is an integer, hence $a \leqslant 1$.
(2) When $a=1$, $bc=1$, the original system of equations can be transformed into
$$
\left\{\begin{array}{l}
k x^{2}+y+(k-1)=0, \\
y=-(k+1) x+1
\end{array}\right.
$$

When $k=0$, the system of equations (1) becomes $\left\{\begin{array}{l}y-1=0, \\ y=-x+1\end{array}\right.$.
The system of equations has a unique solution $x=0, y=1$.
When $k \neq 0$, the system of equations (1) can be transformed into
$$
k x^{2}-(k+1) x+k=0 \text {. }
$$

Let's assume the two roots of this equation are $x_{1}, x_{2}$. Then
$$
x_{1}+x_{2}=\frac{k+1}{k}, x_{1} x_{2}=1 \text {. }
$$

Therefore, $x_{1} x_{2}+x_{1}+x_{2}=\frac{k+1}{k}+1$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=\frac{1}{k}+3 \text {. }
$$

Since the system of equations has only one set of integer solutions, it must be that $x_{1}=x_{2}$ and they are integers, thus,
$$
\Delta=(k+1)^{2}-4 k^{2}=0 \text {. }
$$

Solving this, we get $k=1$ or $k=-\frac{1}{3}$ (discard).
When $k=1$, $\left(x_{1}+1\right)\left(x_{2}+1\right)=4$, then $x_{1}=x_{2}=1$.
So, $y=-1$.
The system of equations has only one solution $x=1, y=-1$.
In summary,
When $k=0$, the solution to the system of equations is $x=0, y=1$;
When $k=1$, the solution to the system of equations is $x=1, y=-1$.