6 A country has $n$ airports, served by $k$ airlines, with some airports connected by direct flights (direct flights are bidirectional, meaning one can travel from $A$ to $B$ and from $B$ to $A$). If two airports are not directly connected, they can still be reached via connecting flights.
(1) To connect these $n$ airports, $n-1$ routes are needed (you can use this conclusion without proof). Due to frequent airline bankruptcies, airlines need to closely cooperate in their flight routes. How many direct flight routes must these airlines provide at a minimum to ensure that even if any one airline goes bankrupt, the remaining flight services can still connect any airport to any other airport?
(2) When $n=7, k=5$, to ensure that even if any 2 airlines go bankrupt simultaneously, the remaining flight services can still connect any airport to any other airport, how many direct flight routes must these airlines provide at a minimum?
6. (1) Suppose the $i$-th airline ($i=1,2, \cdots, k$) provides $a_{i}$ direct flights, and let $S=a_{1}+a_{2}+\cdots+a_{k}$. When the $i$-th airline goes bankrupt, the number of direct flights provided by the remaining airlines is $S-a_{i}$. Therefore, $S-a_{i} \geqslant n-1$.

Thus, $\sum_{i=1}^{k}\left(S-a_{i}\right) \geqslant \sum_{i=1}^{k}(n-1)=k(n-1)$, which implies $k S-S \geqslant k(n-1)$, so $S \geqslant \frac{k(n-1)}{k-1}$.

Since $S$ is an integer, $S \geqslant\left[\frac{k(n-1)+k-2}{k-1}\right]=\left[\frac{k n-2}{k-1}\right]$.
Conversely, we prove that $S=\left[\frac{k n-2}{k-1}\right]$ satisfies the conditions, i.e., there exists an $n$-vertex graph $G$ such that $\|G\|=\left[\frac{k n-2}{k-1}\right]$, and the edges of $G$ can be $k$-colored (each edge is colored with one of $k$ colors) such that removing any color of edges (corresponding to the bankruptcy of an airline) leaves the remaining graph connected.

We use induction on $n$ (with a step of $k-1$).
When $n=1,2, \cdots, k-1$, $n<k$, at this time $S=\left[\frac{k n-2}{k-1}\right]=n+\left[\frac{n-2}{k-1}\right]=n$. Construct a cycle of length $n$, and color its edges with $k$ colors such that no two edges have the same color (since $n<k$, this is possible). Thus, removing any color of edges leaves the remaining graph connected, and the conclusion holds.

Assume the conclusion holds for $n \leqslant r$ (where $r \geqslant k-1$), and consider the case $n=r+k-1$.
Take $r$ vertices $A_{1}, A_{2}, \cdots, A_{r}$, by the induction hypothesis, there exists an $n$-vertex graph $G$ with $\left[\frac{k n-2}{k-1}\right]$ edges that can be colored as required, with the $k$ colors being $1,2, \cdots, k$.

Take a vertex $A_{1}$ in $G$, and $k-1$ additional vertices $B_{1}, B_{2}, \cdots, B_{k-1}$ outside $G$, and connect the edges $B_{i-1} B_{i}$ ($i=1,2, \cdots, k$, where $B_{0}=B_{k}=A_{1}$), to form the graph $G^{\prime}$. Then $\left\|G^{\prime}\right\|=\|G\|+k=\left[\frac{k n-2}{k-1}\right]+k=\left[\frac{k(n+k-1)-2}{k-1}\right]$, and color the edge $B_{i-1} B_{i}$ with the $i$-th color. We prove that this coloring satisfies the requirements.

Indeed, suppose we remove the edges of color $i$ ($i=1,2, \cdots, k$), and consider any two vertices $A$ and $B$ in $G^{\prime}$.

If $A, B \in \mathrm{V}(G)$, then by the induction hypothesis, $A$ and $B$ are connected;
If $A, B \in\left\{B_{1}, B_{2}, \cdots, B_{k-1}\right\}$, then since $B_{0}=A_{1}, B_{1}, B_{2}, \cdots, B_{k-1}$ form a cycle of length $k$, with each color appearing exactly once, removing one edge of a color leaves $A$ and $B$ connected;

If $A \in \mathrm{V}(G)$ and $B \in\left\{B_{1}, B_{2}, \cdots, B_{k-1}\right\}$, then $A=A_{1}$, or by the induction hypothesis, $A$ is connected to $A_{1}$, and $B$ is connected to $A_{1}$ through the cycle $\left(A_{1}, B_{1}, B_{2}, \cdots, B_{k-1}\right)$, thus $A$ and $B$ are connected.

Therefore, the conclusion holds for $n=r+k-1$, so these airlines together provide at least $\left[\frac{k n-2}{k-1}\right]$ direct flights.
(2) When $n=7$ and $k=5$, if a certain airport is connected by only 2 direct flights, then when the airlines providing these two flights go bankrupt, the airport becomes unreachable, so each airport must be connected by at least 3 direct flights, thus there are at least $3 \cdot 7=21$ direct flights. Since each direct flight is counted twice (once for each airport), $S \geqslant \frac{21}{2}$, but $S$ is an integer, so $S \geqslant 11$.
When $S=11$, let the 5 airlines be labeled as 1, $2,3,4,5$, then the 11 direct flights shown in the figure below satisfy the requirements.

Therefore, all these airlines together provide at least 11 direct flights.