= Solution
The intersection calculation in part (vi)(a) gives $F^2=0$. Part (vi)(b), with its necessary $F\ne0$ qualification, excludes every nonzero fixed part. Thus $F=0$ and $|D|$ is movable. Part (v) gives $\dim|D|\geq1$, and $D^2=0$ by hypothesis. Applying part (ii) proves
$$
\boxed{|D|\text{ is basepoint-free},\qquad\kappa(D)=1.}
$$
Thus a nontrivial <nef divisor> of square zero on a <K3 surface> defines a fibration over a curve; the key step is <fixed-part elimination on a K3 surface>.
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