Prove or disprove: for any comm. ring $R $,
For three $R $-modules $M $, $N $ and $P $, $M \oplus P \simeq N \oplus P $ implies $M \simeq N $.
Let $f:M\to N $ and $g:N\to M $ be two $R $-module homomorphisms such that $gf=1_M $ (where $1_M = \operatorname{id}_M$ means the identity map on $M$). Then, $N\simeq\operatorname{Im}(f)\oplus\ker(g)$.
If $R $ is a PID and the modules are f.g., 1 is true, but I have no idea on general cases.
I think $1$ is false, even when $R$ is a PID. Let $R=\mathbb{Z}$, and $M=\mathbb{Z}$, $N=\mathbb{Z}\oplus \mathbb{Z}$, and $P=\oplus_{n=1}^{\infty} \mathbb{Z}$. Then $M \oplus P \simeq N \oplus P$ as they are both isomorphic to $\oplus_{n=1}^{\infty} \mathbb{Z}$, but $M\not\simeq N$ as $\mathbb{Z}$-modules (they have different rank).
For part (2), I see that Jyrki has a hint in the comment, but let me say few words. This fact is known as splitting lemma in some algebra books. The fact that $g\circ f = 1_{M}$ implies that $g$ is surjective and $f$ is injective. Hence, we have an exact sequence: $$ 0 \longrightarrow M\overset{f}\longrightarrow N\overset{g}\longrightarrow M \longrightarrow 0 $$ One of the criteria for the exact sequence to split is to have a map $h: M\to N$ such that $g\circ h = 1_{M}$. In this case, $h=f$ fulfills this.