= Solution
An <orthogonal matrix> satisfies $A^{-1}=A^T$, while a <symmetric matrix> satisfies $A^T=A$. Combining them gives $\boxed{A^{-1}=A}$. For the product with the <antisymmetric matrix> $B$, the rule for the <transpose> of a <matrix product> gives
$$
(AB)^T=B^TA^T=-BA.
$$
Thus $AB$ is an <antisymmetric matrix> exactly when $AB=BA$. The hypotheses do not force this <commutativity>. For a counterexample, take
$$
A=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad AB=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
Here $A$ is a <symmetric matrix> and an <orthogonal matrix>, and $B$ is an <antisymmetric matrix> and an <orthogonal matrix>, but $AB$ is a nonzero <symmetric matrix>, so it is not an <antisymmetric matrix>. On the other hand, the product of two <orthogonal matrices> is always an <orthogonal matrix>:
$$
(AB)^T(AB)=B^TA^TAB=B^TB=I.
$$
Therefore the answers are $\boxed{\text{(i) no; (ii) yes.}}$
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