1. The range of the function $y=\frac{x^{2}+2 x+3}{x^{2}+4 x+5}(x \in \mathbf{R})$ is
,$- 1 . y \in[2-\sqrt{2}, 2+\sqrt{2}]$.
From the problem, we know
$$
y=\frac{x^{2}+2 x+3}{x^{2}+4 x+5}
$$

is equivalent to the equation in terms of $x$
$$
(y-1) x^{2}+(4 y-2) x+(5 y-3)=0
$$

having real roots, which means
$$
\begin{array}{l}
(2 y-1)^{2}-(y-1)(5 y-3) \geqslant 0 \\
\Rightarrow y^{2}-4 y+2 \leqslant 0 \\
\Rightarrow 2-\sqrt{2} \leqslant y \leqslant 2+\sqrt{2} .
\end{array}
$$