1. Given the sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the relation $\left(2-a_{n+1}\right)\left(4+a_{n}\right)=8$, and $a_{0}=2$, then $\sum_{i=0}^{n} \frac{1}{a_{i}}=$
$\qquad$ . $\qquad$
$$
\begin{array}{l}
2 a_{n}-4 a_{n+1}=a_{n} a_{n+1} \Rightarrow \frac{2}{a_{n+1}}-\frac{4}{a_{n}}=1 \Rightarrow \frac{1}{a_{n+1}}=\frac{2}{a_{n}}+\frac{1}{2} \\
\Rightarrow \frac{1}{a_{n+1}}+\frac{1}{2}=2\left(\frac{1}{a_{n}}+\frac{1}{2}\right), \text{ and } \frac{1}{a_{0}}=\frac{1}{2} \text{, then } \frac{1}{a_{n}}+\frac{1}{2}=2^{n} \Rightarrow \frac{1}{a_{n}}=2^{n}-\frac{1}{2} .
\end{array}
$$

Therefore, $\sum_{i=0}^{n} \frac{1}{a_{i}}=2^{n+1}-1-\frac{n+1}{2}=\frac{1}{2}\left(2^{n+2}-n-3\right)(n \in \mathbf{N})$.