= Solution
The <quotient theorem for Cartesian tensors> here says: if contracting $T_{ij}$ with every <vector> $v_j$ gives the components $w_i=T_{ij}v_j$ of a <vector> in every right-handed <orthonormal basis>, then $T$ obeys the second-order <tensor> transformation law. Conversely that transformation law guarantees the contraction is a <vector>. One may equivalently test that $u_iT_{ij}v_j$ is a <scalar> for every pair of <vectors> $u,v$.
By part (a), $v'=Rv$ and $w'=Rw$. The contraction property in the new <basis> says
$$
T'Rv=w'=Rw=RTv.
$$
Since this holds for every test <vector>, $T'R=RT$. Using $R^{-1}=R^T$ gives
$$
\boxed{T'=RTR^T,\qquad T'_{ij}=R_{ik}R_{j\ell}T_{k\ell}.}
$$
For the converse, $T'v'=RTR^TRv=RTv=Rw$, exactly the <vector> law. For the equivalent <scalar> formulation, invariance of $u^TTv$ for every $u,v$ gives $u^TR^TT'Rv=u^TTv$, forcing the same <matrix> identity.
The quantifier “every” matters: an arbitrary nonzero <matrix> can annihilate one fixed <vector>, so a single successful contraction cannot establish tensoriality. The stated <bases> test proper rotations; transformation under <orthogonal reflections> would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the <matrix rank> of a particular component <matrix>.
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