Five. (20 points) Given $a, b > 0$. Find the maximum values of
$$
\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}}, \sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{b+2 a}}
$$
$$
\begin{array}{l}
\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}} \\
\leqslant \sqrt{2\left(\frac{a}{2 a+b}+\frac{b}{2 b+a}\right)} .
\end{array}
$$

Let $a+b=1$.
Then the right side of equation (1) $=\sqrt{2\left(\frac{a}{1+a}+\frac{b}{1+b}\right)}$
$$
=\sqrt{4-2\left(\frac{1}{1+a}+\frac{1}{1+b}\right)} \text {. }
$$

Since $\left(\frac{1}{1+a}+\frac{1}{1+b}\right)(1+a+1+b) \geqslant 4$,
i.e., $\frac{1}{1+a}+\frac{1}{1+b} \geqslant \frac{4}{3}$,
thus, $\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}} \leqslant \frac{2 \sqrt{3}}{3}$.
Equality holds if and only if $a=b$.
Let $x=\frac{a}{b}$, then the original problem can be transformed into finding the maximum value of the function $y=\sqrt{\frac{x}{x+2}}+\sqrt{\frac{1}{1+2 x}}(x>0)$.
$y^{\prime}=\sqrt{\frac{x+2}{x}} \cdot \frac{1}{(x+2)^{2}}-\frac{\sqrt{1+2 x}}{(1+2 x)^{2}}$.
Let $y^{\prime}=0$. After rearranging, we get $(x+1)(x-1)^{3}=0$, obtaining the critical point $x=1$.
When $0<x<1$, $f^{\prime}(x)>0$;
When $x>1$, $f^{\prime}(x)<0$.
Therefore, the maximum value of the function $y=\sqrt{\frac{x}{x+2}}+\sqrt{\frac{1}{1+2 x}}(x>0)$ is $\frac{2 \sqrt{3}}{3}$.