Let $\triangle A B C$ have internal angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively,
vector $\boldsymbol{m}=(\sin A, b+c)$,
$$
\boldsymbol{n}=(\sin C-\sin B, a-b),
$$

and there exists a real number $\lambda$ such that $\boldsymbol{m}=\lambda \boldsymbol{n}$.
(1) Find the size of $\angle C$;
(2) If $a+b=k c$, find the range of the real number $k$.
(1) From $\boldsymbol{m}=\lambda \boldsymbol{n}$, we get
$$
\left\{\begin{array}{l}
\sin A=\lambda(\sin C-\sin B), \\
b+c=\lambda(a-b) .
\end{array}\right.
$$

Eliminating $\lambda$ yields
$$
(a-b) \sin A=(b+c)(\sin C-\sin B).
$$

By the Law of Sines, we have
$$
(a-b) a=(b+c)(c-b),
$$

which simplifies to $a^{2}+b^{2}-c^{2}=a b$. Therefore, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{1}{2}$.
Since $0<\angle C<\pi$, we have $\angle C=\frac{\pi}{3}$.
(2) From the conclusion in (1), we get
$$
\angle B=\frac{2 \pi}{3}-\angle A.
$$

By the given information and the Law of Sines, we have
$$
\begin{aligned}
k & =\frac{a+b}{c}=\frac{\sin A+\sin B}{\sin C} \\
& =\frac{\sin A+\sin \left(\frac{2 \pi}{3}-A\right)}{\sin \frac{\pi}{3}} \\
& =\sqrt{3} \sin A+\cos A=2 \sin \left(A+\frac{\pi}{6}\right).
\end{aligned}
$$

Since $0<\angle A<\frac{2 \pi}{3} \Rightarrow \frac{\pi}{6}<\angle A+\frac{\pi}{6}<\frac{5 \pi}{6}$
$$
\Rightarrow \frac{1}{2}<\sin \left(A+\frac{\pi}{6}\right) \leqslant 1.
$$

Thus, the range of $k$ is $(1,2]$.