= Solution
Use <Dynkin labels> $w_i=\langle w,\alpha_i^\vee\rangle$, so the rows of the <Cartan matrix> are the <simple roots> expressed in <fundamental weight> coordinates. For the <A2 root system>,
$$
\boxed{A_{A_2}=\begin{pmatrix}2&-1\\-1&2\end{pmatrix},\qquad
\alpha_1=(2,-1),\quad\alpha_2=(-1,2).}
$$
Starting at the <highest weight> $(1,0)$ and subtracting successive allowed <simple roots> gives $(1,0)\to(-1,1)\to(0,-1)$. Negating these <weights> gives the <dual representation>. Hence the two fundamental <weight systems> are
$$
\boxed{\begin{aligned}
\mathbf3=(1,0):&\quad (1,0),\ (-1,1),\ (0,-1),\\
\overline{\mathbf3}=(0,1):&\quad (0,1),\ (1,-1),\ (-1,0).
\end{aligned}}
$$
Each listed <weight> has <weight multiplicity> one. The <highest weight> $(1,1)=\alpha_1+\alpha_2$ is the <highest root>, so its <representation> is the <Adjoint representation>. Its nonzero <weights> are the six <roots>, while the <Cartan subalgebra> supplies the zero <weight> twice:
$$
\boxed{\mathbf8=(1,1):\quad
\pm(2,-1),\quad\pm(-1,2),\quad\pm(1,1),\quad(0,0)^{[2]}.}
$$
The superscript $[2]$ indicates <weight multiplicity>. This also follows from $\mathbf3\otimes\overline{\mathbf3}=\mathbf8\oplus\mathbf1$: the nine <tensor product> <weights> comprise six differences of distinct fundamental <weights> and three zeros, one zero belonging to the <trivial representation>.
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