Let $N=\{1,2, \cdots, n\}, n \geqslant 2 . N$'s subsets $A_{i}(i=1,2, \cdots, t)$ form the set $F=\left\{A_{1}\right.$, $\left.A_{2}, \cdots, A_{t}\right\}$. If for every pair of elements $x, y \in \mathbf{N}$, there is a set $A_{i} \in F$ such that $A_{i} \cap\{x, y\}$ contains exactly one element, then $F$ is called separable. If every element of $N$ belongs to at least one set $A_{i} \in F$, then $F$ is called covering. What is the minimum value $f(n)$ of $t$ such that there exists an $F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ that is both separable and covering?
Let $F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ be both separable and covering for $N$. Consider the table representing the relationship between sets and elements:

\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{l} 
Element \\
Set
\end{tabular} & 1 & 2 & 3 & $\ldots$ & $n$ \\
\hline$A_{1}$ & $a_{11}$ & $a_{12}$ & $a_{13}$ & $\cdots$ & $a_{1 n}$ \\
\hline$A_{2}$ & $a_{21}$ & $a_{22}$ & $a_{23}$ & $\ldots$ & $a_{2 n}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$A_{t}$ & $a_{t 1}$ & $a_{t 2}$ & $a_{t 3}$ & $\cdots$ & $a_{t n}$ \\
\hline
\end{tabular}

where $a_{i j}=\left\{\begin{array}{l}1, j \in A \\ 0, j \notin A\end{array} \quad 1 \leqslant i \leqslant t, 1 \leqslant j \leqslant n\right.$.
(1) Since $F$ is covering, each $j$ belongs to at least one $A_{i}$, i.e., each column in the table contains at least one 1.
(2) Since $F$ is separable, no two columns in the table are identical.

Since each element in the $t$ rows can only be 0 or 1, and no column is entirely 0, the maximum number of distinct columns that can be formed is $2^{t}-1$. By the separability of $F$ (or by (2)), we have
$$
n \leqslant 2^{t}-1 \implies \log _{2} n \geqslant\left[\log _{2} n\right] \text {, thus } \quad f(n) \geqslant\left[\log _{2} n\right]+1 \text {. (1) }
$$

On the other hand, let $t$ satisfy $2^{i-1} \leqslant n \leqslant 2^{i}-1$, i.e., $t=\left[\log _{2} n\right]+1$.
We can construct $n$ distinct columns of length $t$ consisting of 0s and 1s, and none of these columns are entirely 0, because $n \leqslant 2^{t}-1$, this is always possible. We can use these columns as the $n$ columns of a $t \times n$ table, and interpret this table as the relationship between sets $A_{1}, A_{2}, \cdots, A_{t}$ and elements $1,2, \cdots, n$. Specifically, the set $A_{i}$ consists of the $j$ values for which $a_{i j}=1$, i.e., $A_{i}=\left\{j \mid 1 \leqslant j \leqslant n\right.$ and $\left.a_{i j}=1\right\}, 1 \leqslant i \leqslant t$.
In this case, the set $F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ is both separable and covering for $N$, so we also have
$$
f(n) \leqslant t=\left[\log _{2} n\right]+1 \text {. (2) }
$$

From (1) and (2), we conclude that $f(n)=\left[\log _{2} n\right]+1$.