= Solution
For the <D5 root system>, take an orthonormal <basis> $e_1,\ldots,e_5$ and <simple roots> $\alpha_i=e_i-e_{i+1}$ for $i=1,2,3,4$, together with $\alpha_5=e_4+e_5$. The <Dynkin diagram> has single bonds $1-2$, $2-3$, $3-4$, $3-5$, with the fork at node $3$. Thus its <Cartan matrix> is
$$
\boxed{C_{D_5}=\begin{pmatrix}
2&-1&0&0&0\\
-1&2&-1&0&0\\
0&-1&2&-1&-1\\
0&0&-1&2&0\\
0&0&-1&0&2
\end{pmatrix}.}
$$
The <fundamental weight> $\omega_5=\tfrac12(e_1+e_2+e_3+e_4+e_5)$ has <Dynkin labels> $(0,0,0,0,1)$. Its pairing with every positive <coroot> $e_i\pm e_j$, $i<j$, is zero or one. Hence its <half-spin representation> is a <minuscule representation>, so its <weight system> is a single <Weyl group> orbit with multiplicity one. The <Weyl group> of the <D5 root system> permutes coordinates and changes an even number of signs. Consequently its complete <weight system> is
$$
\boxed{\mu=\tfrac12(s_1e_1+s_2e_2+s_3e_3+s_4e_4+s_5e_5),
\quad s_i\in\{1,-1\},\quad\prod_i s_i=1.}
$$
There are $1+\binom52+\binom54=16$ such <weights>. Their <Dynkin labels> are $((s_1-s_2)/2,(s_2-s_3)/2,(s_3-s_4)/2,(s_4-s_5)/2,(s_4+s_5)/2)$, explicitly
$$
\begin{array}{rr}
(0,0,0,0,1)&(0,0,1,0,-1)\\
(0,1,-1,1,0)&(0,1,0,-1,0)\\
(1,-1,0,1,0)&(1,-1,1,-1,0)\\
(1,0,-1,0,1)&(1,0,0,0,-1)\\
(-1,0,0,1,0)&(-1,0,1,-1,0)\\
(-1,1,-1,0,1)&(-1,1,0,0,-1)\\
(0,-1,0,0,1)&(0,-1,1,0,-1)\\
(0,0,-1,1,0)&(0,0,0,-1,0)
\end{array}
$$
Here the <Lie algebra> is $\mathfrak{so}(10,\mathbb C)$, or its compact real form when discussing compact <groups>. Globally the <half-spin representation> is a <representation> of $\operatorname{Spin}(10)$, and does not descend to $SO(10)$; the question only specifies the <Lie algebra>.
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