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 $\mathbf{Q}([n]^{-1}(E(\mathbf{Q})))$, 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 $n$. 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 $\mathbf{Q}(\phi^{-1}(E(\mathbf{Q})))$.

My questions are in the opposite direction:

1. Do there exist infinitely many $p$ such that $\mathrm{red}_p$ is surjective?

2. Do there exist arbitrarily large sets of primes $\{ p_1, p_2, \ldots, p_m \}$ such the combined reduction map $E(\mathbf{Q}) \rightarrow \prod_{i=1}^m \widetilde{E}(\mathbf{F}_{p_i})$ is surjective?

3. 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 as sketched in the first paragraph?