14. (15 points) Given positive real numbers $x, y$ satisfy
$$
\begin{array}{l}
a=x+y, \\
b=\sqrt{x^{2}+7 x y+y^{2}} .
\end{array}
$$
(1) When $y=1$, find the range of $\frac{b}{a}$;
(2) If $c^{2}=k x y$, and for any positive numbers $x, y$, the segments with lengths $a, b, c$ can always form a triangle, find the range of $k$.
14. (1) Notice,
$$
\begin{array}{l}
\frac{b}{a}=\frac{\sqrt{x^{2}+7 x+1}}{x+1} \\
=\sqrt{\frac{x^{2}+7 x+1}{x^{2}+2 x+1}} \\
=\sqrt{1+\frac{5 x}{x^{2}+2 x+1}} \\
=\sqrt{1+\frac{5}{x+2+\frac{1}{x}}} .
\end{array}
$$

Given $x+2+\frac{1}{x} \geqslant 4$, we know
$$
0\sqrt{x^{2}+7 x y+y^{2}}-(x+y),\end{array}\right.$ which means
$$
\left\{\begin{array}{l}
\sqrt{k}\frac{\sqrt{x^{2}+7 x y+y^{2}}-(x+y)}{\sqrt{x y}}
\end{array}\right.
$$

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

always holds.
$$
\text { Let } \frac{x}{y}=t \text {. Then } t \in(0,+\infty) \text {. }
$$

Since $t+\frac{1}{t}$ is an increasing function for $t \in[1,+\infty)$, we know
$$
f(t)=\sqrt{t+\frac{1}{t}+7}+\sqrt{t+\frac{1}{t}+2}
$$

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

Notice,
$$
\begin{array}{l}
\sqrt{t+\frac{1}{t}+7}-\sqrt{t+\frac{1}{t}+2} \\
=\frac{5}{\sqrt{t+\frac{1}{t}+7}+\sqrt{t+\frac{1}{t}+2}}
\end{array}
$$
$\leqslant 1$.
Therefore, $1<\sqrt{k}<5$, which means $1<k<25$.
Thus, the range of $k$ is $(1,25)$.