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False. The Carmichael number $561=3\cdot 11\cdot 17$ supplies a counterexample. For every $a$ coprime to $561$, the Fermat little theorem gives $a^{560}\equiv 1$ modulo each of these primes, since $2,10,16$ all divide $560$. The Chinese remainder theorem then gives $a^{560}\equiv 1(\mathrm{mod}561)$. Thus the proposed Fermat primality test can fail to detect a composite for every coprime base.

True. First suppose $p^e\mid N$ with $e\ge 2$ maximal. By the Chinese remainder theorem, choose $a\equiv 1+p^{e-1}(\mathrm{mod}{p}^e)$ and $a\equiv 1$ modulo all other prime-power factors. This is a unit, with Jacobi symbol one, because it is one modulo every prime. For $m=(N-1)/2$, the binomial theorem givesbecause $p\nmid m$ and all higher terms contain $p^e$. It is therefore a witness to the desired failure.

If $N$ is square-free and composite, it has at least two prime factors. Choose $a$ to be a quadratic nonresidue at one prime $p$ and one at every other prime, again using the Chinese remainder theorem. Then $(a/N)=-1$. At a different odd prime $q\mid N$, however, $a^m\equiv 1\not\equiv -1(\mathrm{mod}q)$. These two cases prove that every odd composite has an Euler-Jacobi witness, unlike the situation for the ordinary Fermat primality test.