$\newcommand{\Sha}{Ш}\newcommand{\alg}{\mathrm{alg}}\DeclareMathOperator\Sel{Sel}\DeclareMathOperator\rank{rank}$ Consider the elliptic curves $E$ of $j$-invariant zero that neither them nor their isogenous curves have any rational torsion points. From the $3$-descent sequence on these curves we have that the algebraic rank \begin{equation}(1) \quad \rank_{\alg}(E/\mathbb{Q}) = \dim_{\mathbb{F}_{3}}(E(\mathbb{Q})/3E(\mathbb{Q})) = \dim_{\mathbb{F}_{3}}\Sel_{3}(E/\mathbb{Q}) - \dim_{\mathbb{F}_{3}}\Sha(E/\mathbb{Q})[3].\end{equation}
If we assume finiteness of the $3$-primary part $\Sha(E/\mathbb{Q})[3^{\infty}]$, then from the Cassels-Tate pairing we have that $\Sha(E/\mathbb{Q})[3]$ has even dimension hence from (1) above we get that $\rank_{\alg}(E/\mathbb{Q}) = \dim_{\mathbb{F}_{3}}(E(\mathbb{Q})/3E(\mathbb{Q}))$ has the same parity as $\dim_{\mathbb{F}_{3}}\Sel_{3}(E/\mathbb{Q})$. Let us call this the parity result.
Question 1: Are there any examples of such elliptic curves (or any other elliptic curves not necessarily of $j$-invariant zero and not necessarily for $p=3$) that have rank $> 1$ where the parity result is proved without assuming finiteness of $\Sha(E/\mathbb{Q})[3^{\infty}]$?
Question 2: If one shows the parity result for an elliptic curve without assuming finiteness of $\Sha(E/\mathbb{Q})[3^{\infty}]$, would that imply the finiteness of $\Sha(E/\mathbb{Q})[3^{\infty}]$?
Question3: Has the finiteness of $\Sha(E/\mathbb{Q})[3^{\infty}]$ been proved for any elliptic curve of rank $>1$?