The integers $\mathbb{Z}$ are a principal ideal domain. Hence by this answer we know that any submodule of a free module is free. Thus we have a resolution of free Abelian groups:
$$\cdots\to 0\to 0\to 0\to F_1 \stackrel{d_0}\to \mathbb{Z}^\mathbb{R}\stackrel{\epsilon}\dashrightarrow\mathbb{R}\to 0,$$
where $\epsilon(e_\lambda)=\lambda$.
Notation: I will slightly abuse notation to write:
$$H_i(M; N)={\rm Tor}_i^{\mathbb{Z}}(M,N),\qquad H^i(M; N)={\rm Ext}^i_{\mathbb{Z}}(M,N)$$
Clearly from this resolution we have $H_0(\mathbb{R};\mathbb{Z})=\mathbb{R}$ and $H_1(\mathbb{R};\mathbb{Z})=0$.
However we can have nontrivial first cohomology groups:
$$H^1(\mathbb{R};F_1)\neq 0.$$
Proof:. Consider the identity map $1_{F_1}\colon F_1\to F_1$. This is an element of $H^1(\mathbb{R};F_1)$ with respect to the above resolution. If it were trivial (as an element of cohomology), then the map $d_0$ would split. Hence $\epsilon$ would split.
Let $\delta\colon \mathbb{R}\to \mathbb{Z}^\mathbb{R}$ be such a splitting map. Write $\delta(1)=\sum_{i=1}^k n_i e_{\lambda_i}$ for some integers $n_i\neq0$ and distinct real numbers $\lambda_i$. Note we must have $k>0$ or $\epsilon\delta(1)=0\neq 1$, contradicting $\epsilon\delta=1_\mathbb{R}$.
Then there is no value that $\delta\left(\frac 1{2n_1}\right)$ could take, so that $\delta$ is an abelian group homomorphism. We conclude that $1_F$ represents a non-trivial element of cohomology. $\qquad\qquad$
$\Box$
Update: From the fact that $F_1=\oplus_{i\in I}\mathbb{Z}$, for some indexing set $I$, it is tempting to conclude that $$H^1(\mathbb{R};\mathbb{Z})\neq 0.$$ However this does not follow in any obvious way that I can see. The problem is that for abelian groups $A_i$, we have ${\rm Hom}(\bullet,\oplus_{i\in I}A_i)$ lies in a no-man's land between better behaved functors: $$\bigoplus_{i\in I}{\rm Hom}(\bullet,A_i)
\subseteq
{\rm Hom}(\bullet,\bigoplus_{i\in I}A_i)
\subseteq \prod_{i\in I}{\rm Hom}(\bullet,A_i).$$
Thus it is not clear how to relate $H^1(\mathbb{R};F_1)=H^1(\mathbb{R};\oplus_{i\in I}\mathbb{Z})$ to $H^1(\mathbb{R};\mathbb{Z})$.
However we can prove $H^1(\mathbb{R};\mathbb{Z})\neq 0$ a more long-winded way:
Proof: We have $\mathbb{R}=\oplus_{i\in I} \mathbb{Q}$ for some indexing set $I$. Thus $$H^1(\mathbb{R};\mathbb{Z})=\prod_{i\in I}H^1(\mathbb{Q};\mathbb{Z}).$$
Then it suffices to show that $H^1(\mathbb{Q};\mathbb{Z})\neq 0$.
We have a resolution:
$$\cdots\to 0\to 0\to 0\to \mathbb{Z}^{\mathbb{N}_{>0}} \stackrel{d_0}\to \mathbb{Z}^{\mathbb{N}_\geq 0}\stackrel{\epsilon}\dashrightarrow\mathbb{Q}\to 0,$$
where $\epsilon(e_i)=\frac 1{i!}$ and $d_0(f_i)=ie_i-e_{i-1}$.
Then a $1$-cocycle $x$ will map each $f_i\mapsto x_i$ for a sequence of integers $x_i$.
This cocycle will be a coboundary precisely when we have a sequence of integers $y_i$, so that the map $y\colon e_i\mapsto y_i$ satisfies $d_0^*(y)=x$. That is $$x_i=iy_i-y_{i-1}.$$
This can be rearranged into a recurrence relation:$$y_i=\frac{y_{i-1}+x_i}i.$$
For any initial value $y_0$, the above recurrence relation determines a sequence of rational numbers $y_i$. Then $x$ is a coboundary precisely when for some integer $y_0$, the resulting sequence lies in $\mathbb{Z}$.
Well-order the integers. Let $x_1=0$ and choose $x_i$ for $i=2,3,4,\cdots$ in turn, so that if $y_0$ is the $(i-1)$'nth integer, then $y_i\notin \mathbb{Z}$. Note that $y_{i-1}$ is determined by the chosen value of $y_0$ and the already established values $x_1,x_2,\cdots,x_{i-1}$. We just need to choose $x_i$ so that $x_i\not\equiv -y_{i-1} \mod i$.
The result of this diagonal argument is a cocycle $x$, such that there is no integer $y_0$, for which the sequence $y_i$ remains in $\mathbb{Z}$. Thus $x$ is not a coboundary and represents a non-trivial element in $H^1(\mathbb{Q};\mathbb{Z})$. Then the element of $H^1(\mathbb{R};\mathbb{Z})$ whose projection to each factor $H^1(\mathbb{Q};\mathbb{Z})$ is $x$, is also non-trivial. $\qquad \qquad \Box$
