Please, if you were kind enough to review this exercise and comment if there is anything to correct.
EJERCICIO 2 Prove that $R$-homomorphism $F : M \rightarrow N$ is a monomorphism if and only if $\ker \left( f \right) = 0$
SOLUCION $\ker\left( f \right) = \left\lbrace x \in M / f \left\lbrace x \right) = 0 \right\rbrace $
$\boxed{\Rightarrow}$
Definition: $f$ is injective $\rightarrow \forall \; a, b \in M : f \left( a \right) = f \left( b \right) \Rightarrow a = b$
Thus choose $a, b \in M$ such that $f \left( a \right) = f \left( b \right) $ then:$$\begin{array}{clc}&f \left( a \right) - f \left( b \right) &= 0\\ \implies &f \left( a - b \right) &= 0\\ \implies &f \left( 0 \right) &= 0\end{array}$$
$\boxed{\Leftarrow}$
Consider $a, b \in M$ and suppose that $f \left( a \right) = f \left( b \right) $. Then $f \left( a - b \right) = f \left( a \right) - f \left( b \right)$ and thus $a - b \in \ker \left( f \right) $. Therefore, the hypothesis $\ker \left( f \right) = \left\lbrace 0 \right\rbrace $ implies that $a - b = 0$ or equivalently $a = b$. This says that $f$ is injective.
Your second part is correct. However your first part is in fact the same as the second one, so that is clearly not correct. You should start from a statement of the form
$$\text{''Consider $a\in M$ such that $f(a)=0$''}$$
and then the first implication should be clear from the definition of injectivity.