The answer to Q2 for $n=3$ is actually 'no, without some nondegeneracy hypotheses'. The reason is as follows:

The curvature tensor $\mathcal{R}= R_{ijkl}\,(\mathrm{d}x^i\wedge\mathrm{d}x^j)\circ(\mathrm{d}x^k\wedge\mathrm{d}x^l)$, with all its indices lowered, is a section of the subbundle $K(M)\subset S^2\bigl(\Lambda^2(T^*M)\bigr)$ that is the kernel of the natural linear mapping
$$
S^2\bigl(\Lambda^2(T^*M)\bigr)\longrightarrow \Lambda^4(T^*M).
$$
You are asking, for a given section $\mathcal{R}$, whether there exists a metric $g$ such that $\mathrm{Riem}(g) = \mathcal{R}$. There is no pointwise algebraic condition on the section $\mathcal{R}$ imposed by this equation (that was what Q1 was about), but, as you note, there is the *second Bianchi identity*
$$
\mu(\nabla^g\mathcal{R}) = 0,
$$
where $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is the natural skewsymmetrization operation. Since $\nabla^g$ depends on one derivative of the metric $g$, the above equation with a given $\mathcal{R}$ can be regarded as a first-order system of equations on $g$. When $n=3$, this is at most $3$ equations, the rank of the bundle $\Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$.

Now, for a point $p\in M$ satisfying $\mathcal{R}(p)=0$, the value $\nabla^g\mathcal{R}(p)$ does not depend on $g$ (just look at the formula for $\nabla^g\mathcal{R}$ in local coordinates). Thus, if $\mathcal{R}(p)=0$, but $\mu(\nabla^{g_0}\mathcal{R})(p) \not= 0$ for some metric $g_0$, then $\mu\bigl(\nabla^{g}\mathcal{R}(p)\bigr) \not= 0$ for all metrics $g$ and hence there is no open neighborhood of $p$ on which the equations $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ have a solution $g$. In particular, the original system $\mathrm{Riem}(g) = \mathcal{R}$ has no solution in a neighborhood of such a point $p$.

To convince yourself that such examples exist when $n=3$, just note that the rank of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$ in this case is $6$, so an arbitrary section that vanishes at $p$ will have $6\times 3 = 18$ independent first derivatives. Thus, the map $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is surjective (and the rank of the target bundle is $3$), so that the generic section of $K(M)$ that vanishes at $p$ will not satisfy the second Bianchi identity at $p$ for any metric $g$.

However, suppose that $n=3$ and that $\mathcal{R}$ is a *nondegenerate* section of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$. I proved (back in the early 1980s) that, when $\mathcal{R}$ is real-analytic, there always exist local solutions to the equation $\mathrm{Riem}(g) = \mathcal{R}$. Specifically, I showed that, in this case, in addition to the $6$ second-order equations that these equations represent on $g$ and the $3$ first-order equations on $g$ that $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ represents, there is one more first-order equation $Q_\mathcal{R}(g)=0$ on $g$ that is satisfied by any metric $g$ that satisfies $\mathrm{Riem}(g) = \mathcal{R}$. Then I proved that the combined overdetermined system of $6$ second-order equations and $4$ first-order equations for $g$ is *involutive*, so that an application of the Cartan-Kähler Theorem proves local solvability. Unfortunately, the involutive system is never either hyperbolic or elliptic, though it can be of real principal type.

I never published my proof, but later, Dennis DeTurck and Deane Yang studied the overdetermined system that I wrote down and published a proof of its local solvability in the smooth category. See Deturck and Yang, *Local existence of smooth metrics with prescribed curvature*, Nonlinear problems in geometry (Mobile, Ala., 1985), 37–43,
Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986.