1. Xét cặp cạnh AC và B'D'.

Ta có $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.

$\overrightarrow{B'D'} = \overrightarrow{B'A'} + \overrightarrow{A'D'} = \overrightarrow{BA} + \overrightarrow{AD} = \overrightarrow{AD} - \overrightarrow{AB}$

$\overrightarrow{AC}.\overrightarrow{B'D'} = (\overrightarrow{AB} + \overrightarrow{BC}).(\overrightarrow{AD} - \overrightarrow{AB}) = \overrightarrow{AB}.\overrightarrow{AD} - \overrightarrow{AB}.\overrightarrow{AB} + \overrightarrow{BC}.\overrightarrow{AD} - \overrightarrow{BC}.\overrightarrow{AB}$

Vì ABCD.A'B'C'D' là hình hộp có các cạnh bằng nhau nên ABCD là hình vuông.

$\overrightarrow{AB}.\overrightarrow{AD} = 0$

$\overrightarrow{AB}.\overrightarrow{AB} = a^2$

$\overrightarrow{BC}.\overrightarrow{AD} = \overrightarrow{AD}.\overrightarrow{AD} = a^2$

$\overrightarrow{BC}.\overrightarrow{AB} = 0$

$\overrightarrow{AC}.\overrightarrow{B'D'} = 0 - a^2 + a^2 - 0 = 0$

Vậy $AC \perp B'D'$.

2. Xét cặp cạnh AB' và CD'.

$\overrightarrow{AB'} = \overrightarrow{AB} + \overrightarrow{BB'}$

$\overrightarrow{CD'} = \overrightarrow{CD} + \overrightarrow{DD'} = \overrightarrow{BA} + \overrightarrow{AA'} = -\overrightarrow{AB} + \overrightarrow{AA'}$

$\overrightarrow{AB'}.\overrightarrow{CD'} = (\overrightarrow{AB} + \overrightarrow{BB'}).(-\overrightarrow{AB} + \overrightarrow{AA'}) = -\overrightarrow{AB}.\overrightarrow{AB} + \overrightarrow{AB}.\overrightarrow{AA'} - \overrightarrow{BB'}.\overrightarrow{AB} + \overrightarrow{BB'}.\overrightarrow{AA'}$

$= -a^2 + 0 - 0 + a^2 = 0$

Vậy $AB' \perp CD'$.

3. Xét cặp cạnh AD' và CB'.

$\overrightarrow{AD'} = \overrightarrow{AD} + \overrightarrow{DD'} = \overrightarrow{AD} + \overrightarrow{AA'}$

$\overrightarrow{CB'} = \overrightarrow{CB} + \overrightarrow{BB'} = -\overrightarrow{BC} + \overrightarrow{AA'}$

$\overrightarrow{AD'}.\overrightarrow{CB'} = (\overrightarrow{AD} + \overrightarrow{AA'}).(-\overrightarrow{BC} + \overrightarrow{AA'}) = -\overrightarrow{AD}.\overrightarrow{BC} + \overrightarrow{AD}.\overrightarrow{AA'} - \overrightarrow{AA'}.\overrightarrow{BC} + \overrightarrow{AA'}.\overrightarrow{AA'}$

$= -a^2 + 0 - 0 + a^2 = 0$

Vậy $AD' \perp CB'$.

`=>` Đpcm