The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively.

In fact, $A$ is amenable if and only if $(A\hat{\otimes}_{\pi}A)^{*} = A^{*} \oplus K$ as a direct sum of $A$-bimodules for some sub-$A$-bimodule $K$ of $(A\hat{\otimes}_{\pi}A)^{*}$, see https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s2-40.1.89. From this, it is not difficult to show that $$\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*}) = \mathcal{Z}(A,A^{*}) \oplus\mathcal{Z}(A,K)$$ as a direct sum of $Z(A)$-modules.

It is also not difficult to show that every $\beta\in \mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ is given by $$\beta(x\otimes y) = f(yx) \hspace{8mm} \forall x,y\in A$$ for some $f\in A^{*}$, and every $\beta\in\mathcal{Z}(A,A^{*})$ is similarly given by $f\in A^{*}$ with the extra property that $af=fa$ for every $a\in A$. Hence, one may identify $\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ with $A^{*}$, and $\mathcal{Z}(A,A^{*})$ by $C^{\perp}$ in a very natural way. This, in fact, is the simple observation behind the two formulations above of the same problem.

The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively.

Notation: $A\hat{\otimes}_{\pi} A$ is the projective tensor product of $A$ with itself. For an $A$-bimodule $M$, the bimodule center is defined by $\mathcal{Z}(A,M) =\{\beta\in M: a\beta=\beta a\hspace{4mm} \forall a\in A \}$. $A^{*}$ is identified with the image of the map (dual of the product map) $\sim:A^{*}\to (A\hat{\otimes}_{\pi} A)^{*}$, which is defined by $\widetilde{f}(x\otimes y) = f(xy)$ on basic tensors (and extended linearly & continuously afterwards).

$A$ is amenable if and only if $(A\hat{\otimes}_{\pi}A)^{*} = A^{*} \oplus K$ as a direct sum of $A$-bimodules for some sub-$A$-bimodule $K$ of $(A\hat{\otimes}_{\pi}A)^{*}$, see Curtis & Loy  . From this, it is not difficult to show that $$\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*}) = \mathcal{Z}(A,A^{*}) \oplus\mathcal{Z}(A,K)$$ as a direct sum of $Z(A)$-modules. It is also not difficult to show that every $\beta\in \mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ is given by $$\beta(x\otimes y) = f(yx) \hspace{8mm} \forall x,y\in A$$ for some $f\in A^{*}$, and every $\beta\in\mathcal{Z}(A,A^{*})$ is similarly given by $f\in A^{*}$ with the extra property that $af=fa$ for every $a\in A$. Hence, one may identify $\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ with $A^{*}$, and $\mathcal{Z}(A,A^{*})$ by $C^{\perp}$ in a natural way via $Z(A)$-module isomorphisms. Similarly, $\mathcal{Z}(A,K)$ is isomorphic to a $Z(A)$-submodule of $A^{*}$, call it $B$. Consequently, $$A^{*} = C^{\perp}\oplus B$$ as a direct sum of $Z(A)$ submodules. In particular, $C^{\perp}$ is a complemented subspace of $A^{*}$.

N.B.: Amenability is a considerably strong condition compared to the ones given by Q1 & Q2. Thus, it is interesting to see weaker sufficient conditions. On the contrary, Q1/Q2 provides relatively easy to check tests for non-amenability. Thus, it is interesting to see a class of Banach algebras where Q1 or Q2 is not satisfied.

The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively.

In fact, $A$ is amenable if and only if $(A\hat{\otimes}_{\pi}A)^{*} = A^{*} \oplus K$ as a direct sum of $A$-bimodules for some sub-$A$-bimodule $K$ of $(A\hat{\otimes}_{\pi}A)^{*}$, see https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s2-40.1.89. From this, it is not difficult to show that $$\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*}) = \mathcal{Z}(A,A^{*}) \oplus\mathcal{Z}(A,K)$$ as a direct sum of $Z(A)$-modules.

The purpose of this note is not to answer my own question, but to share a sufficient condition with everyone: If $A$ is amenable, then Q1 & Q2 are answered affirmatively.

In fact, $A$ is amenable if and only if $(A\hat{\otimes}_{\pi}A)^{*} = A^{*} \oplus K$ as a direct sum of $A$-bimodules for some sub-$A$-bimodule $K$ of $(A\hat{\otimes}_{\pi}A)^{*}$, see https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s2-40.1.89. From this, it is not difficult to show that $$\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*}) = \mathcal{Z}(A,A^{*}) \oplus\mathcal{Z}(A,K)$$ as a direct sum of $Z(A)$-modules.

It is also not difficult to show that every $\beta\in \mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ is given by $$\beta(x\otimes y) = f(yx) \hspace{8mm} \forall x,y\in A$$ for some $f\in A^{*}$, and every $\beta\in\mathcal{Z}(A,A^{*})$ is similarly given by $f\in A^{*}$ with the extra property that $af=fa$ for every $a\in A$. Hence, one may identify $\mathcal{Z}(A,(A\hat{\otimes}_{\pi}A)^{*})$ with $A^{*}$, and $\mathcal{Z}(A,A^{*})$ by $C^{\perp}$ in a very natural way. This, in fact, is the simple observation behind the two formulations above of the same problem.