= Solution
For the <special unitary group> $SU(N)$ define $\varepsilon_i=e_i-\frac1N\sum_{j=1}^Ne_j$. These are the <weights of a representation> in the defining space, with $\sum_i\varepsilon_i=0$. Their <Dynkin labels> are
$$
(w_i)_r=(\varepsilon_i,\alpha_r)=\delta_{ir}-\delta_{i,r+1},\qquad1\leq r<N.
$$
The <dual representation> negates each <weight of a representation>, and the <tensor product of group representations> adds them. Consequently the complete <weight systems> are
$$
\boxed{\begin{aligned}
\mathcal W(V)&=\{\varepsilon_i:1\leq i\leq N\},&
\mathcal W(V^*)&=\{-\varepsilon_i:1\leq i\leq N\},\\
\mathcal W(\Lambda^2V)&=\{\varepsilon_i+\varepsilon_j:1\leq i<j\leq N\},&
\mathcal W(\Lambda^2V^*)&=\{-\varepsilon_i-\varepsilon_j:1\leq i<j\leq N\}.
\end{aligned}}
$$
Every listed <weight of a representation> has multiplicity one: the corresponding <weight basis> vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in <Dynkin labels>, the two defining <weight systems> are
$$
\begin{array}{c|c|c}
i&A_5:\mathbf6&A_4:\mathbf5\\\hline
1&(1,0,0,0,0)&(1,0,0,0)\\
2&(-1,1,0,0,0)&(-1,1,0,0)\\
3&(0,-1,1,0,0)&(0,-1,1,0)\\
4&(0,0,-1,1,0)&(0,0,-1,1)\\
5&(0,0,0,-1,1)&(0,0,0,-1)\\
6&(0,0,0,0,-1)&\text{not present}
\end{array}
$$
The <weight systems> for $\overline{\mathbf6}$ and $\overline{\mathbf5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.
For the <exterior squares>, the full enumerations are
$$
\begin{array}{c|c|c}
ij&A_5:\mathbf{15}&A_4:\mathbf{10}\\\hline
12&(0,1,0,0,0)&(0,1,0,0)\\
13&(1,-1,1,0,0)&(1,-1,1,0)\\
14&(1,0,-1,1,0)&(1,0,-1,1)\\
15&(1,0,0,-1,1)&(1,0,0,-1)\\
16&(1,0,0,0,-1)&\text{not present}\\
23&(-1,0,1,0,0)&(-1,0,1,0)\\
24&(-1,1,-1,1,0)&(-1,1,-1,1)\\
25&(-1,1,0,-1,1)&(-1,1,0,-1)\\
26&(-1,1,0,0,-1)&\text{not present}\\
34&(0,-1,0,1,0)&(0,-1,0,1)\\
35&(0,-1,1,-1,1)&(0,-1,1,-1)\\
36&(0,-1,1,0,-1)&\text{not present}\\
45&(0,0,-1,0,1)&(0,0,-1,0)\\
46&(0,0,-1,1,-1)&\text{not present}\\
56&(0,0,0,-1,0)&\text{not present}
\end{array}
$$
The negatives of the $A_5$ column are exactly the fifteen <weights of a representation> of $\overline{\mathbf{15}}$; the negatives of the $A_4$ column are exactly the ten of $\overline{\mathbf{10}}$. Thus the tables specify all eight <weight systems>, including multiplicities, rather than only their <highest weights>.
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