= Solution
For the standard <A4 root system> inside the <A5 root system>, let $p$ delete the fifth <Dynkin label>. Put $v_i=p(w_i)$ for $i\le5$. These are precisely the five <weights> of the defining $SU(5)$ <fundamental representation>:
$$
(v_1,v_2,v_3,v_4,v_5)=\bigl((1,0,0,0),(-1,1,0,0),(0,-1,1,0),(0,0,-1,1),(0,0,0,-1)\bigr).
$$
Since $p(w_6)=0$, the six projected <weights> of $\overline{\mathbf6}$ are $-v_1,\ldots,-v_5,0$. The first five form the <dual representation> $\overline{\mathbf5}$, with <highest weight> $(0,0,0,1)$; the last is a singlet. Thus
$$
\boxed{\overline{\mathbf6}\downarrow_{SU(5)}=\overline{\mathbf5}\oplus\mathbf1.}
$$
For $\mathbf{15}$, split the fifteen <weights> in <4/b> into the ten pairs with $1\le i<j\le5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete <weight decomposition> of $\Lambda^2\mathbf5=\mathbf{10}$, with <highest weight> $(0,1,0,0)$. The latter project to $v_i$, the complete <weight decomposition> of $\mathbf5$. Therefore
$$
\boxed{\mathbf{15}\downarrow_{SU(5)}=\mathbf{10}\oplus\mathbf5.}
$$
This <restriction of a representation> can also be checked without a table: $V\downarrow_{SU(5)}=\mathbf5\oplus\mathbf1$, so its <dual representation> is $\overline{\mathbf5}\oplus\mathbf1$, while its <exterior square> is $\Lambda^2\mathbf5\oplus(\mathbf5\otimes\mathbf1)$. The dimensions match $6=5+1$ and $15=10+5$.
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