1. Given $A_{1}, A_{2}, \cdots, A_{n}$ are 11 stations sequentially on a straight highway, and
$$
\begin{array}{l}
A_{i} A_{i+2} \leqslant 12(i=1,2, \cdots, 9), \\
A_{i} A_{i+3} \geqslant 17(i=1,2, \cdots, 8) .
\end{array}
$$

If $A_{1} A_{11}=56 \mathrm{~km}$, then $A_{2} A_{7}=$ $\qquad$ $\mathrm{km}$.
Ni, 1.29.
According to the problem, we have $A_{1} A_{10} \geqslant 3 \times 17=51(\mathrm{~km})$, then
$$
A_{10} A_{11} \leqslant 5 \mathrm{~km} \text {; }
$$

Also, $A_{8} A_{11} \geqslant 17 \mathrm{~km}, A_{8} A_{10} \leqslant 12 \mathrm{~km}$, then $A_{10} A_{11} \geqslant 5 \mathrm{~km}$.
Therefore, $A_{10} A_{11}=5 \mathrm{~km}$.
Similarly, $A_{1} A_{2}=5 \mathrm{~km}$.
And $A_{1} A_{7} \geqslant 34 \mathrm{~km}$, thus,
$$
A_{2} A_{7} \geqslant 29 \mathrm{~km} \text {. }
$$

Also, $A_{2} A_{10}=56-5 \times 2=46(\mathrm{~km}), A_{7} A_{10} \geqslant 17 \mathrm{~km}$, so $A_{2} A_{7} \leqslant 29 \mathrm{~km}$.
From equations (1) and (2), we get $A_{2} A_{7}=29 \mathrm{~km}$.