Sure, here is the translated text:

```
(40 points) Let $x, y$ be non-negative real numbers,
$$
a=\sqrt{x}+\sqrt{y}, b=\sqrt{x+2}+\sqrt{y+2} \text{.}
$$

If $a, b$ are two non-consecutive integers, find the values of $a, b$.
```
Since $a$ and $b$ are non-adjacent integers, we have
$$
\begin{array}{l}
2 \leqslant b-a=(\sqrt{x+2}-\sqrt{x})+(\sqrt{y+2}-\sqrt{y}) \\
=\frac{2}{\sqrt{x+2}+\sqrt{x}}+\frac{2}{\sqrt{y+2}+\sqrt{y}} \\
\leqslant \frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}=2 \sqrt{2}<3 .
\end{array}
$$

Since $b-a$ is an integer, it follows that $b-a=2$.
Let $a=n-1, b=n+1 (n \in \mathbf{Z})$, then
$$
\sqrt{x}+\sqrt{y}=n-1, \sqrt{x+2}+\sqrt{y+2}=n+1 \text {. }
$$

Thus, $\sqrt{x}-\sqrt{y}=\frac{x-y}{n-1}$,
$$
\sqrt{x+2}-\sqrt{y+2}=\frac{x-y}{n+1} \text {. }
$$

Therefore, $2 \sqrt{x}=n-1+\frac{x-y}{n-1}$,
$$
2 \sqrt{x+2}=n+1+\frac{x-y}{n+1} \text {. }
$$

Hence, $2(n-1) \sqrt{x}=(n-1)^{2}+(x-y)$, $2(n+1) \sqrt{x+2}=(n+1)^{2}+(x-y)$.
Thus, $(n-1) \sqrt{x}+2 n=(n+1) \sqrt{x+2}$.
$$
\text { Also, }(\sqrt{x+2})^{2}-(\sqrt{x})^{2}=2 \text {, }
$$

Let $t=\sqrt{x}$. Then
$$
\sqrt{x+2}=\frac{(n-1) t+2 n}{n+1} .
$$

Substituting into equation (1) gives
$$
\begin{array}{l}
2 n t^{2}-2 n(n-1) t-\left(n^{2}-2 n-1\right)=0 . \\
\text { Therefore, } \sqrt{x}=t=\frac{n(n-1) \pm(n+1) \sqrt{n(n-2)}}{2 n}, \\
\sqrt{y}=n-1-\sqrt{x} \\
=\frac{n(n-1) \mp(n+1) \sqrt{n(n-2)}}{2 n} .
\end{array}
$$

Thus, $n \geqslant 2$, and
$$
n(n-1) \geqslant(n+1) \sqrt{n(n-2)} \text {, }
$$

which implies $n^{2}-2 n-1 \leqslant 0$.
Solving this, we get $1-\sqrt{2} \leqslant n \leqslant 1+\sqrt{2}$.
Thus, $2 \leqslant n \leqslant 1+\sqrt{2}$, and $n \in \mathbf{Z}$.
Hence, $n=2$.
Therefore, $a=1, b=3$.