= Solution
To compute the <first Tor of a cyclic quotient>, start a <free resolution> of $A/I$ with $P_0=A$, take a free module $P_1$ surjecting onto $I$, and take $P_2$ surjecting onto the kernel of $P_1\to I$. Tensor right exactness identifies
$$
\operatorname{coker}(P_2\otimes_AM\longrightarrow P_1\otimes_AM)
\simeq I\otimes_AM.
$$
By definition, the degree-one <Tor functor> is the kernel of $P_1\otimes_AM\to A\otimes_AM$ modulo the image from $P_2\otimes_AM$. Quotienting first by that image consequently gives the natural identification
$$
\boxed{\operatorname{Tor}_1^A(A/I,M)\simeq
\ker(\mu_I:I\otimes_AM\longrightarrow M).}
$$
Equivalently the calculation gives the exact sequence
$$
0\longrightarrow\operatorname{Tor}_1^A(A/I,M)
\longrightarrow I\otimes_AM\xrightarrow{\mu_I}M
\longrightarrow M/IM\longrightarrow0.
$$
Thus its first term vanishes for every ideal exactly when each multiplication map is injective, proving
$$
\boxed{(3)\Longleftrightarrow(4).}
$$
Together with the preceding implications, \b[all four conditions are equivalent]. The proof used explicit extension arguments, character duality and the definition of the <Tor functor>, rather than an unproved flatness or injectivity criterion.
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