 Solution

For the D5 root system, take an orthonormal basis $e_1,\dots ,e_5$ 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 isThe fundamental weight ${\omega }_5=\frac{1}{2}(e_1+e_2+e_3+e_4+e_5)$ 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 isThere are $1+\left(\genfrac{}{}{0ex}{}{5}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{4}\right)=16$ such weights. Their Dynkin labels are $((s_1-s_2)/2,(s_2-s_3)/2,(s_3-s_4)/2,(s_4-s_5)/2,(s_4+s_5)/2)$, explicitlyHere 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 $\mathrm{Spin}(10)$, and does not descend to $SO(10)$; the question only specifies the Lie algebra.