For $4 \Rightarrow 1$ assume $P$ satisfies $4$ and let $M \twoheadrightarrow N$ be surjective with a map $P \to N$. The pullback $M \times_N P \to P$ is surjective (you can prove this categorically or from the definition of the object using that $M \twoheadrightarrow N$ is surjective. Alternatively, see An Introduction to Homological Algebra by Rotman, Exercise 5.10 on page 227) so by $4$ it is just a projection to a summand isomorphic to $P$, say $M\times_N P \cong P\oplus Q$. Then the inclusion of that summand, and the other half of the pullback square ($P \to P\oplus Q \cong M \times_N P \to M \to N$) gives that $P \to N$ factors through $M$. Thus $P$ is projective (I assume you're using the definition stating that $P$ is projective if given an epimorphism $ψ:M→N$ and any morphism $f:P→N$, there exists a morphism (not necessarily unique) $g:P→M$ such that $f=ψ\circ g$).
I suspect a dual argument using the pushforward works for $4' \Rightarrow 1'$, but I haven't thought about it.
Edit: Note that $i_P\colon P \to M \times_N P$ splits $\pi_2$, so $\pi_2\circ i_P = \mathrm{id}_P$. Then $f = f\circ\pi_2\circ i_P = \phi\circ\pi_1\circ i_P$ so $f$ factors through $\phi$.
Theorem. Every (right) module is a submodule of an injective module.
The proof of this will be given later on, for the moment assume it. Using right modules is easier, but the result of course holds also for left modules.
Conditions 1 and 2 are clearly equivalent, as are conditions 3 and 4.
Suppose 2 holds. Let $Q'$ be an injective module such that $Q\subseteq Q'$. Let $j$ be the inclusion map. By the injectivity of $Q'$, the homomorphism $ig\colon X\to Q'$ can be extended to a homomorphism $h'\colon Y\to Q'$ such that $h'f=jg$. Since the inclusion $j$ splits, let $q\colon Q'\to Q$ be such that $qj=1_Q$ (the identity on $Q$). Then
$$
qh'f=qjg=g
$$
and $h=qh'\colon Y\to Q$ is the homomorphism we were looking for.
Assume 3 holds and let $Q$ be a submodule of $M$. Then the inclusion map $j\colon Q\to M$ admits an extension $q\colon M\to M$ such that $qj=1_Q$, so $Q$ is a direct summand of $M$.
Therefore 2 and 3 are equivalent and we are done.
The existence of an embedding $M\to E$ where $E$ is injective can be proved in the following way. The module $M$ embeds into $\operatorname{Hom}_{\mathbb{Z}}(R,M)$. If $M$, considered as an abelian group, is embedded in the abelian group $N$, then $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ is embedded in $\operatorname{Hom}_{\mathbb{Z}}(R,N)$, as it's easy to show. So we just need two lemmas.
Lemma. If $N$ is a divisible group, then $\operatorname{Hom}_{\mathbb{Z}}(R,N)$ is an injective $R$-module.
Proof. Let $f\colon X\to Y$ be an injective homomorphism and suppose $g\colon X\to \operatorname{Hom}_{\mathbb{Z}}(R,N)$ is a module homomorphism. Then we can consider $g'\colon X\to N$ defined by $g'(x)=g(x)(1)$. This is a group homomorphism, so by the divisibility of $N$, there is a group homomorphism $h'\colon Y\to N$ such that $h'f=g'$. Now define $h\colon Y\to\operatorname{Hom}_{\mathbb{Z}}(R,N)$ by $h(y)(r)=h'(yr)$.
Then, for $x\in X$ and $r\in R$,
$$
hf(x)(r)=h(f(x))(r)=h'(f(x)r)=h'(f(xr))=g'(xr)=g(xr)(1)=g(x)(r)
$$
so $hf=g$. QED
Note: the $R$-module structure on $\operatorname{Hom}_{\mathbb{Z}}(R,N)$ is given by defining $fr$ to be the map $fr\colon s\mapsto f(rs)$.
Lemma. Every abelian group embeds in a divisible group.
Proof. If $G$ is an abelian group, it is not restrictive to see $G=F/K$, where $F$ is a free abelian group. A free abelian group is a direct sum of copies of $\mathbb{Z}$, that so embeds in a direct sum $F'$ of copies of $\mathbb{Q}$, which is divisible. Then $G$ embeds in $F'/K$, which is also divisible. QED
Finally, notice that for abelian groups (or $\mathbb{Z}$-modules), being injective is equivalent to being divisible (apply Baer's criterion).
Best Answer
Since $$\text{Hom}_A(M,\text{Hom}_{\Bbb Z}(A,\Bbb R/\Bbb Z))\cong\text{Hom}_{\Bbb Z} (M\otimes_A A,\Bbb R/\Bbb Z)\cong\text{Hom}_{\Bbb Z}(M,\Bbb R/\Bbb Z)$$ so what one needs is an Abelian group homomorphism $g:M\to\Bbb R/\Bbb Z$ with $g(x)\ne0$. There is a nonzero homomorphism from $\Bbb Z x$ to $\Bbb R/\Bbb Z$ and we can extend this by a Zorn's lemma argument to an Abelian group homomorphism $g:M\to \Bbb R/\Bbb Z$.
If you want, you may be more explicit about how this defines a homomorphism $f:M\to\text{Hom}_{\Bbb Z}(A,\Bbb R/\Bbb Z)$. We let $f(m)$ be the map $a\mapsto g(ma)$ from $A$ to $\Bbb R/\Bbb Z$.