Step 1: Statement (I) via explicit construction: for any \(c \in [-1,1]\), pick \(t^* \in (1, 1+2\pi)\) with \(\sin(t^*) = c\), possible because \((1, 1+2\pi)\) already spans one complete period of \(\sin\), so \(\sin\) restricted here is onto \([-1,1]\). Setting \(x^* = 1/t^* \in (0,1)\) gives \(\sin(1/x^*) = c\). So every real number in \([-1,1]\) is attained, and the set has the cardinality of the continuum.
Step 2: Statement (II) via an explicit uncountable subset: the map \(t \mapsto (t, 1/t)\) from \((0,\infty)\) to \(\mathbb{R}^2\) is injective, so its image has the cardinality of the continuum. Every point \((t, 1/t)\) satisfies \(xy = 1 \in \mathbb{Z}\), so this image is a subset of \(\{(x,y): xy \in \mathbb{Z}\}\). A set containing an uncountable subset is itself uncountable.
Step 3: Statement (III) via a different family: take \(B_b = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}\) for \(b \in \mathbb{R}\). Its eigenvalues are both \(1\), which is rational, regardless of \(b\), and distinct values of \(b\) give distinct matrices, so this again produces an uncountable set of matrices with rational eigenvalues.
Step 4: All three sets are uncountable, so (I), (II) and (III) are all true.
\[\boxed{\text{All (I), (II) and (III) are true}}\]