(14) Given positive real numbers $x, y$, let $a=x+y, b=\sqrt{x^{2}+7 x y+y^{2}}$.
(1) When $y=1$, find the range of $\frac{b}{a}$;
(2) If $c^{2}=k \cdot x y$, and for any positive numbers $x, y$, the segments of lengths $a, b, c$ can always form a triangle, find the range of $k$.


14 (1) Method 1 From the given, we know that $x=a-1$, and $a=x+1>1$. Therefore,
$$
\begin{aligned}
\frac{b}{a} & =\frac{\sqrt{(a-1)^{2}+7(a-1)+1}}{a}=\sqrt{\frac{a^{2}+5 a-5}{a^{2}}} \\
& =\sqrt{-5\left(\frac{1}{a}-\frac{1}{2}\right)^{2}+\frac{9}{4}},
\end{aligned}
$$

Since $a=x+1>1 \Rightarrow \frac{1}{a} \in(0,1)$, combining with the graph of the quadratic function, we know
$$
1\sqrt{x^{2}+7 x y+y^{2}}-(x+y),
\end{array}\right.
$$

which means
$$
\left\{\begin{array}{l}
\sqrt{k}\frac{\sqrt{x^{2}+7 x y+y^{2}}-(x+y)}{\sqrt{x y}}
\end{array}\right.
$$

always holds, i.e.,
$$
\left\{\begin{array}{l}
\sqrt{k}\sqrt{\frac{x}{y}+\frac{y}{x}+7}-\sqrt{\frac{x}{y}+\frac{y}{x}+2}
\end{array}\right.
$$

always holds.
Let $\frac{x}{y}=t$, then $t \in(0,+\infty)$, since $t+\frac{1}{t}$ is an increasing function on $t \in[1,+\infty)$, then
$$
f(t)=\sqrt{t+\frac{1}{t}+7}+\sqrt{t+\frac{1}{t}+2}
$$

is an increasing function on $[1,+\infty)$. Therefore,
$$
f(t)=\sqrt{t+\frac{1}{t}+7}+\sqrt{t+\frac{1}{t}+2} \geqslant \sqrt{9}+\sqrt{4}=5 .
$$

Also,
$$
\sqrt{t+\frac{1}{t}+7}-\sqrt{t+\frac{1}{t}+2}=\frac{5}{\sqrt{t+\frac{1}{t}+7}+\sqrt{t+\frac{1}{t}+2}} \leqslant 1 \text {, }
$$

so $1<\sqrt{k}<5,1<k<25$, i.e., the range of $k$ is $(1,25)$.