1. Given that $a$ and $b$ are both unit vectors, and vector $c=a+2 b$ is perpendicular to $d=5 a-4 b$. Then the angle between $a$ and $b$ $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=$ $\qquad$
$$
-1 \cdot \frac{\pi}{3} \text {. }
$$

Let the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\langle\boldsymbol{a}, \boldsymbol{b}\rangle=\theta$. Then, according to the condition for vectors to be perpendicular, we have
$$
\begin{array}{c}
0=c \cdot d=(a+2 b) \cdot(5 a-4 b) \\
=5 a^{2}+10 a \cdot b-4 a \cdot b-8 b^{2} \\
=5+6 \cos \theta-8=6 \cos \theta-3 .
\end{array}
$$

Thus, $\cos \theta=\frac{1}{2}$. Therefore, $\theta=\frac{\pi}{3}$.