Let $M$ be $R$-module and $N$ be submodule of $M$. Let $x,y\in M$. Prove or disprove:
(1) If $x,y\in N$ then $x+y\in N$.
(2) If $x\in N$ and $y\not\in N$ then $x+y\not\in N$.
(3) If $x\not\in N$ and $y\not\in N$ then $x+y\in N$.
I try as below
(1) Obviously, since $N$ be submodule of $M$ then $x+y\in N$.
(2) I doubt it. I take example $M=\mathbb{Z}$:$\mathbb{Z}$-module and the submodule is $N=\mathbb{2Z}$:$\mathbb{Z}$-module. Now, take $2n+1\not\in \mathbb{2Z}$ and $2m\in \mathbb{2Z}$, $m,n\in\mathbb{Z}$. \begin{align} (2n+1)+2m=4n+1=2(n+m)+1\not\in\mathbb{2Z}. \end{align} Generally, does $x+y\not\in N$ holds?
(3) I doubt it. I take example $M=\mathbb{Z}$:$\mathbb{Z}$-module and the submodule is $N=\mathbb{2Z}$:$\mathbb{Z}$-module. Now, take $2n+1\not\in \mathbb{2Z}$ and $2m+1\not\in \mathbb{2Z}$, $m,n\in\mathbb{Z}$. \begin{align} (2n+1)+(2m+1)=2(n+m)+2=2((n+m)+1)\in\mathbb{2Z}. \end{align} Generally, does $x+y\in N$ holds?
(1) is certainly correct.
For (2) you gave an example, but you can prove it in general. Indeed, if $x + y \in N$ then $y = (x + y) - x \in N$, a contradiction.
For (3) your example doesn't make sense. You found $x, y \notin N$ and showed $x + y \in N$, but since you said you doubted it you want to show exactly the opposite. Here's an example: take the $\mathbb Z$ module $M = \mathbb Z \times \mathbb Z$ and let $N = \{(n, 0) : n \in \mathbb Z\}$. Then $(0, 1), (0, 2) \notin N$ and $(0, 1) + (0, 2) = (0, 3) \notin N$.