1. Given $k$ is an irrational number, $b d f>0$,
$$
\frac{a-c}{b-d}=\frac{c-e}{d-f}=\frac{e-k a}{f-k b}=\frac{\sqrt{2}}{2} \text {. }
$$

Then the value of $\frac{a^{2}+c^{2}+e^{2}+a c f+b c e+a d e}{b^{2}+d^{2}+f^{2}+3 b d f}$ is
$=1 . \frac{1}{2}$.
Since $b d f>0$, it follows that $b \neq 0, d \neq 0, f \neq 0$.
Also, $k$ is an irrational number, so $k \neq 1, 1-k \neq 0$. By the property of proportions, we have
$$
\frac{(a-c)+(c-e)+(e-k a)}{(b-d)+(d-f)+(f-k b)}=\frac{\sqrt{2}}{2},
$$

which simplifies to $\frac{a-k a}{b-k b}=\frac{a(1-k)}{b(1-k)}=\frac{\sqrt{2}}{2}$.
Therefore, $\frac{a}{b}=\frac{\sqrt{2}}{2}$.
Thus, $\frac{a-c}{b-d}=\frac{c-e}{d-f}=\frac{a}{b}=\frac{\sqrt{2}}{2}$.
By the property of proportions, we know
$$
\begin{array}{l}
\frac{c}{d}=\frac{e}{f}=\frac{a}{b}=\frac{\sqrt{2}}{2} \\
\Rightarrow \frac{a^{2}}{b^{2}}=\frac{c^{2}}{d^{2}}=\frac{e^{2}}{f^{2}}=\frac{a c f}{b d f}=\frac{b c e}{b d f}=\frac{a d e}{b d f}=\frac{1}{2} .
\end{array}
$$

By the property of proportions, we have
$$
\frac{a^{2}+c^{2}+e^{2}+a c f+b c e+a d e}{b^{2}+d^{2}+f^{2}+3 b d f}=\frac{1}{2} \text {. }
$$