Three. (50 points) Given $a, b, \lambda \in \mathbf{R}_{+}$. Determine the range of
$$
g(a, b)=\sqrt{\frac{a}{a+\lambda b}}+\sqrt{\frac{b}{b+\lambda a}}
$$
Three, by replacing $a, b$ with $\frac{a}{a+b}, \frac{b}{a+b}$ respectively, $g(a, b)$ remains unchanged. Therefore, under the condition $a+b=1$, find the range of $g(a, b)$.
$$
\begin{array}{l}
g^{2}(a, b)=\frac{a}{a+\lambda b}+\frac{b}{b+\lambda a}+2 \sqrt{\frac{a b}{(a+\lambda b)(b+\lambda a)}} \\
=\frac{\lambda+2(1-\lambda) a b}{\lambda+(1-\lambda)^{2} a b}+2 \sqrt{\frac{a b}{\lambda+(1-\lambda)^{2} a b}} .
\end{array}
$$

When $\lambda \neq 1$,
$$
g^{2}(a, b)
=\frac{1}{1-\lambda}\left[2-\frac{\lambda(1+\lambda)}{\lambda+(1-\lambda)^{2} a b}\right]+\frac{2}{11-\lambda} \sqrt{1-\frac{\lambda}{\lambda+(1-\lambda)^{2} a b}}.
$$
Let $t=\sqrt{1-\frac{\lambda}{\lambda+(1-\lambda)^{2} a b}}$.
Since $a+b=1$, then $a b \in\left(0, \frac{1}{4}\right]$.
Thus, $t \in\left(0, \frac{|\lambda-1|}{1+\lambda}\right]$.
Let $f(t)=\frac{2}{1-\lambda}-\frac{1+\lambda}{1-\lambda}\left(1-t^{2}\right)+\frac{2}{|1-\lambda|} \cdot t$.
$$
=\frac{1+\lambda}{1-\lambda} \cdot t^{2}+\frac{2}{|1-\lambda|} \cdot t+1 \text {. }
$$

When $0<\lambda<1$,
$$
f(t)=\frac{1+\lambda}{1-\lambda}\left(t-\frac{1}{1+\lambda}\right)^{2}+\frac{\lambda^{2}}{\lambda^{2}-1}, t \in\left(0, \frac{\lambda-1}{\lambda+1}\right] .
$$
Thus,
$$
f(0)<f(t) \leqslant f\left(\frac{\lambda-1}{\lambda+1}\right),
$$

i.e., $1<f(t) \leqslant \frac{4}{1+\lambda}$.
(2) When $2 \leqslant \lambda<3$, combining with the graph, we have
$$
f(0)<f(t) \leqslant f\left(\frac{1}{1+\lambda}\right),
$$

i.e., $1<f(t) \leqslant \frac{\lambda^{2}}{\lambda^{2}-1}$.
(3) When $\lambda \geqslant 3$, combining with the graph, we have
$$
f\left(\frac{\lambda-1}{\lambda+1}\right) \leqslant f(t) \leqslant f\left(\frac{1}{1+\lambda}\right) \text {, }
$$

i.e., $\frac{4}{1+\lambda} \leqslant f(t) \leqslant \frac{\lambda^{2}}{\lambda^{2}-1}$.
When $\lambda=1$, it is clear that
$$
1<f(t) \leqslant 2=\frac{4}{1+\lambda} \text {. }
$$

In summary, we have
When $0<\lambda<2$, $1<g(a, b) \leqslant \frac{2}{\sqrt{1+\lambda}}$;
When $2 \leqslant \lambda<3$, $1<g(a, b) \leqslant \frac{\lambda}{\sqrt{\lambda^{2}-1}}$;
When $\lambda \geqslant 3$, $\frac{2}{\sqrt{1+\lambda}} \leqslant g(a, b) \leqslant \frac{\lambda}{\sqrt{\lambda^{2}-1}}$.
Equality $g(a, b)=\frac{2}{\sqrt{1+\lambda}}$ holds if and only if $a=b$;
Equality $g(a, b)=\frac{\lambda}{\sqrt{\lambda^{2}-1}}$ holds if and only if $\frac{a}{b}=\frac{\lambda^{3}-3 \lambda \pm\left(\lambda^{2}-1\right) \sqrt{\lambda^{2}-4}}{2}$.