= Solution
For a <convex cone> $D$ in an <inner product> space, use the nonnegative-pairing convention
$$
D^*=\{B:\langle A,B\rangle\geq0\text{ for every }A\in D\}.
$$
Here the pairing is the <Frobenius inner product> on real <symmetric matrices>. Let $C_0$ be the <conic hull> of the nonnegative <rank-one matrices> $xx^T$, and let $C=\overline{C_0}$. For $A\in K$, $\langle A,xx^T\rangle_F=x^TAx\geq0$. This extends to <conic combinations>, and by <continuity> to their limits. Hence $C\subseteq K^*$.
For the converse, if $B\notin C$, <separation from a closed convex cone> supplies a symmetric $H$ with
$$
\langle H,B\rangle_F<0,\qquad
\langle H,Z\rangle_F\geq0\quad(Z\in C).
$$
In particular $x^THx\geq0$ for every $x\geq0$, so $H\in K$. The negative pairing then excludes $B$ from $K^*$. Therefore
$$
\boxed{K^*=\overline{\operatorname{cone}\{xx^T:x\geq0\}}}.
$$
The same generator test gives $C^*=K$. This is the <duality of copositive and completely positive cones>; the next argument removes the closure.
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