Here is a full answer to Question 1 in the special case $S=X_1+X_2$. I give an exact upper and lower bounds for $\mathbb P(S\ge x)$.

As mentioned in the comments, for $x\ge 0$ and $\mathbb P(S\le x)$ may be 1
(with a similar result for $\mathbb P(S\ge x)$ when $x\le 0$ by symmetry):
if $X_2$ is taken to be $-X_1$, then $\mathbb P(S\le x)$ is 1 for all $x>0$.

The interesting remaining case is then $\mathbb P(S\ge x)$ when $x>0$ (or its symmetric version $\mathbb P(S\le x)$ when $x<0$). In this case, we show
$\mathbb P(S\ge x)$ is $2\mathbb P(N\ge \frac x2)$.

To see this, fix $x>0$ and define a pair of random variables as follows:

Let $(Z_1,Z_2)$ be $(t,x-t)$ with one-dimensional probability density $f_{N_1}(t)$ for $t\in [\frac x2,\infty)$.

Let $(Z_1,Z_2)$ be $(x-t,t)$ with one-dimensional probability density $f_{N_2}(t)$ for $t\in [\frac x2,\infty)$.

Let $(Z_1,Z_2)$ be $(-\infty,-\infty)$ with the remaining probability.

Now for $t\ge \frac x2$, we can check that $\mathbb P(Z_1\ge t)=\mathbb P(N\ge t)$ and similarly with $Z_2$. Also $\mathbb P(Z_1\ge t)\le \mathbb P(N\ge t)$ for each $t<\frac x2$. In particular, we have $\mathbb P(Z_1\ge t),\mathbb P(Z_2\ge t)\le \mathbb P(N\ge t)$ for each $t$.

We can now define $(X_1,X_2)$ to be $(Z_1,Z_2)$ when the pair is finite, and to "fill in" the remaining probability (only on pairs with both coordinates less than $\frac x2$) to have the correct marginals. We see that $\mathbb P(X_1+X_2\ge x)=\mathbb P(X_1+X_2=x)=2\mathbb P(N\ge \frac x2)$.

Hence it is possible for $\mathbb P(X_1+X_2\ge x)$ to be as large as $2\mathbb P(N\ge \frac x2)$. On the other hand, $\{X_1+X_2\ge x\}\subset\{X_1\ge \frac x2\}\cup\{X_2\ge\frac x2\}$ so that $\mathbb P(X_1+X_2\ge x)\le \mathbb P(X_1\ge \frac x2)+\mathbb P(X_2\ge \frac x2)\le 2\mathbb P(N\ge \frac x2)$, giving a matching upper bound.

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