= Solution
The unique <symmetric and antisymmetric parts of a matrix> are $S=(M+M^T)/2$ and $A=(M-M^T)/2$, giving
$$
\boxed{S=\frac12\begin{pmatrix}0&a+b\\a+b&0\end{pmatrix},\qquad A=\frac12\begin{pmatrix}0&b-a\\a-b&0\end{pmatrix}.}
$$
Their <matrix products> with their <transposes> are $S^TS=(a+b)^2I/4$ and $A^TA=(a-b)^2I/4$. Each is an <orthogonal matrix> precisely when the corresponding factor is one. Thus $a+b=\pm2$ and $a-b=\pm2$, with independent sign choices, and the simultaneous possibilities are
$$
\boxed{(a,b)=(2,0),\ (-2,0),\ (0,2),\ (0,-2).}
$$
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