= Solution
Use the Hermitian $H_a$ from <2/a> to record real <weight> coordinates: $H_a e_j=(w_j)_a e_j$. Reading off the diagonal entries gives
$$
\boxed{w_1=(1,0,0),\quad w_2=(-1,1,0),\quad
w_3=(0,-1,1),\quad w_4=(0,0,-1).}
$$
These are <weights> in the simple-<coroot> coordinate system, not coordinates in an orthonormal <basis>. In the skew-Hermitian convention they mean $(iH_a)e_j=i(w_j)_ae_j$. Intrinsically, the <weight> on $i\operatorname{diag}(h_1,h_2,h_3,h_4)$ is the functional $h\mapsto h_j$, with $\sum_jh_j=0$.
The <conjugate representation> acts by $U^*$, so a diagonal transformation with entries $e^{ih_j}$ becomes one with entries $e^{-ih_j}$. Thus its four <weights> are
$$
\boxed{\bar w_1=(-1,0,0),\quad\bar w_2=(1,-1,0),\quad
\bar w_3=(0,1,-1),\quad\bar w_4=(0,0,1).}
$$
All have multiplicity one. In both <representations> the sum of the <weights> is zero, consistent with the vanishing <trace> of every <Lie algebra generator>.
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