Let $E/\mathbf{Q}$ be an elliptic curve of rank $>0$. It is easy to see that there is a positive-density set of primes $p$ such that the reduction map $\mathrm{red}_p : E(\mathbf{Q}) \rightarrow \widetilde{E}(\mathbf{F}_p)$ is not surjective. Namely, for an integer $n>1$, take any rational prime $p$ that splits completely in the field $K_n = \mathbf{Q}([n]^{-1}(E(\mathbf{Q})))$, i.e. the field that results from adjoining to $\mathbf{Q}$ the coordinates of all preimages of points in $E(\mathbf{Q})$ under multiplication by $n$. Note that the $K_n$ are finite field extensions of $\mathbf{Q}$ (this is equivalent to the weak Mordell-Weil theorem). To show that these $p$ work: take $P \in E(\mathbf{Q})$ and $Q \in E(K_n)$ with $nQ=P$, then $\mathrm{red}_p(P) = n ( \mathrm{red}_p(Q))$, with $\mathrm{red}_p(Q)$ lying in $\widetilde{E}(\mathbf{F}_p)$ by the assumption on $p$, so $\mathrm{red}_p(E(\mathbf{Q}))$ lies in $n \widetilde{E}(\mathbf{F}_p)$, which is an index-$n^2$ subgroup of $\widetilde{E}(\mathbf{F}_p)$.
More generally, for any isogeny $\phi : E' \rightarrow E$ of elliptic curves over $\mathbf{Q}$ with non-trivial kernel, take any prime $p$ that splits completely in the finite field extension $\mathbf{Q}(\phi^{-1}(E(\mathbf{Q})))$ of $\mathbf{Q}$.
My questions are in the opposite direction:
Do there exist infinitely many $p$ such that $\mathrm{red}_p$ is surjective?
Do there exist arbitrarily large sets of primes $\{ p_1, p_2, \ldots, p_m \}$ such that the combined reduction map $E(\mathbf{Q}) \rightarrow \prod_{i=1}^m \widetilde{E}(\mathbf{F}_{p_i})$ is surjective?
For a prime $p$ such that $\mathrm{red}_p$ is not surjective, is the failure of surjectivity explained by some isogeny $\phi$, by the argument sketched in the first paragraph?
