1. Given that $a$, $b$, and $c$ are non-negative real numbers, and $a^{2}+b^{2}+c^{2}+a b+\frac{2}{3} a c+\frac{4}{3} b c=1$. Find the maximum and minimum values of $a+b+c$.
1. First, find the minimum value.

From $(a+b+c)^{2}$
$$
\begin{aligned}
&=a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c \\
&\geqslant a^{2}+b^{2}+c^{2}+a b+\frac{2}{3} a c+\frac{4}{3} b c=1 \\
&\Rightarrow a+b+c \geqslant 1 .
\end{aligned}
$$

Thus, the minimum value of $a+b+c$ is 1, which occurs when one of $a, b, c$ is 1 and the other two are 0.
Next, find the maximum value.
Let $a=a_{1}, b=2 b_{1}, c=3 c_{1}$.
Then the original expression can be transformed into
$$
\begin{aligned}
& a_{1}^{2}+4 b_{1}^{2}+9 c_{1}^{2}+2 a_{1} b_{1}+2 a_{1} c_{1}+8 b_{1} c_{1}=1 . \\
& \text { Let } x_{1}=a_{1}+b_{1}+c_{1}, x_{2}=b_{1}+c_{1}, x_{3}=c_{1} .
\end{aligned}
$$
$$
\begin{aligned}
& \text { Then } a_{1}^{2}+4 b_{1}^{2}+9 c_{1}^{2}+2 a_{1} b_{1}+2 a_{1} c_{1}+8 b_{1} c_{1} \\
&=\left(a_{1}+b_{1}+c_{1}\right)^{2}+3\left(b_{1}+c_{1}\right)^{2}+5 c_{1}^{2} \\
&=x_{1}^{2}+3 x_{2}^{2}+5 x_{3}^{2}=1,
\end{aligned}
$$

and
$$
\begin{aligned}
& a+b+c=a_{1}+2 b_{1}+3 c_{1} \\
&=x_{1}+x_{2}+x_{3} \\
&=1 \cdot x_{1}+\frac{1}{\sqrt{3}} \cdot \sqrt{3} x_{2}+\frac{1}{\sqrt{5}} \cdot \sqrt{5} x_{3} \\
&\leqslant\left(1+\frac{1}{3}+\frac{1}{5}\right)^{\frac{1}{2}}\left(x_{1}^{2}+3 x_{2}^{2}+5 x_{3}^{2}\right)^{\frac{1}{2}} \\
&=\sqrt{\frac{23}{15}}=\frac{\sqrt{345}}{15} .
\end{aligned}
$$

When $x_{1}=M, x_{2}=\frac{1}{3} M, x_{3}=\frac{1}{5} M\left(M=\sqrt{\frac{15}{23}}\right)$, i.e.,
$$
a=\frac{2}{3} M, b=\frac{4}{15} M, c=\frac{3}{5} M
$$

the equality in (1) holds.
In summary, the maximum value of $a+b+c$ is $\frac{\sqrt{345}}{15}$, and the minimum value is 1.