= Solution
A member of both <groups> satisfies $A^TA=I$ and $A^TJA=J$. Multiplication of the second identity by $A$ on the left, using $AA^T=I$, gives $JA=AJ$. A real two-by-two <matrix> commuting with $J=\operatorname{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm1$ by the <orthogonal matrix> identity, and the restriction $a>0$ forces the first to be $1$. Therefore \b[the intersection is the two-element <group>]
$$
\boxed{\{I,J\}=\{\operatorname{diag}(1,1),\operatorname{diag}(1,-1)\}.}
$$
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