= Solution
Use the conventional additive pedigree relatedness $r_{UV}=2\phi_{UV}$, where $\phi_{UV}$ is the <kinship coefficient>, the <probability> that randomly selected gene copies from $U$ and $V$ are <identical by descent>. For noninbred individuals this convention gives ordinary <full siblings> relatedness $1/2$, as required. It is important that this is not the <probability> of sharing at least one copy: that <probability> is $3/4$ for ordinary <full siblings>.
Let the parents be $M,F$. They are otherwise outbred <first cousins>, so $\phi_{MF}=1/16$ and $\phi_{MM}=\phi_{FF}=1/2$. For two children $C,D$, independent <Mendelian segregation> gives
$$
\phi_{CD}=\frac14(\phi_{MM}+\phi_{MF}+\phi_{FM}+\phi_{FF})=\frac9{32}.
$$
Therefore the autosomal additive relatedness is
$$
\boxed{r_{CD}=9/16.}
$$
The additional $1/16$ beyond $1/2$ comes from maternal-paternal cross-origin matches. Each child has <inbreeding coefficient> $1/16$. The additive coefficient is a sum of descent <probabilities> rather than a literal bounded <probability> in all inbred pedigrees; a relationship normalized to self-relatedness would be $r_{CD}/(1+F)=9/17$. Likewise, <identity-by-descent sharing of inbred full siblings> gives a different matched-copy statistic. The exam's calibration to $1/2$ supports the additive convention, but the quoted <probability> wording needs this qualification.
For sex chromosomes the route of cousinship matters, not merely the fact that the parents have opposite sex. Let $P_M$ be the mother-cousin's parent in the shared sibling pair, and $P_F$ the father-cousin's parent in that pair. Let $F_X$ be the <probability> that the mother's randomly transmitted X matches the father's single X by descent. With no founder inbreeding, the possibilities are
$$
\begin{array}{c|c|c}
P_M& P_F&F_X\\\hline
\text{male}&\text{male}&0\\
\text{female}&\text{male}&0\\
\text{male}&\text{female}&1/8\\
\text{female}&\text{female}&3/16
\end{array}
$$
In the first two rows the father-cousin is related through his father, whose X he does not inherit. In the third row his mother is a sister of the mother-cousin's father: the relevant brother and sister match their maternal X copies with <probability> $1/4$, and the mother-cousin transmits her paternal X with <probability> $1/2$, giving $1/8$. In the last row the two sisters' random X gametes match with <probability> $3/8$: their common father's X contributes $1/4$, and their common mother's two X copies contribute $1/8$. Transmission by the mother-cousin halves this to $3/16$. This derives the <sex-linked cousin kinship> values.
For an unweighted comparison by chromosome pair, two daughters have sex-pair additive relatedness $3/4+F_X$: their paternal X always matches, maternal X matches with <probability> $1/2$, and the two cross-origin terms each contribute $F_X$. Two sons have <mean> sex-pair relatedness $3/4$, from an X match of <probability> $1/2$ and a common paternal Y with <probability> one. A daughter-son pair has sex-pair value $1/4+F_X/2$, with the Y contribution zero. These calculations treat chromosomes as unrecombined blocks and weight X and Y equally; chromosome-length-weighted or locus-specific coefficients are different.
For two children with independent equally likely sexes, these three cases have weights $1/4,1/4,1/2$. Their <mean> sex-pair relatedness is $1/2+F_X/2$, so averaging the 22 autosomal pairs and the sex pair gives
$$
\boxed{\overline r_{\rm genome}=\frac{22(9/16)+1/2+F_X/2}{23}.}
$$
If the cousins' route has $F_X=0$, this is $103/184$, slightly below $9/16$. If, as an additional sampling assumption, the four parent-sex routes in the table are equally likely, $\mathbb E F_X=5/64$ and the answer is $1653/2944$. These are different legitimate numerical corrections under different specified pedigrees. \b[There is no unique sex-chromosome correction from opposite-sex cousinhood alone.] The table and formula give it once the missing pedigree and offspring-sex information is supplied.
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