Here is an answer using Tor:
Consider the exact sequence
$$ 0\rightarrow\mathbb{Z}\rightarrow F\rightarrow K\rightarrow 0,$$
where the map from $\mathbb{Z}$ to $F$ sends the generator $1$ to $a$, and $K$ is the cokernel. Tensoring with $H$ we get the long exact sequence
$$ \cdots Tor(F,H)\rightarrow Tor(K,H)\rightarrow H\rightarrow F\otimes H\rightarrow H\otimes K\rightarrow 0. $$
Now $Tor(F,H)=0$ since $F$ is torsion-free. So the kernel of $H\rightarrow F\otimes H$, given by sending $h$ to $a\otimes h$, is $Tor(K,H)$. This is not always non-zero, and the problem as stated is incorrect. Consider $H=\mathbb{Z}/2\mathbb{Z}$, $F=\mathbb{Z}$, $a=2$.
Here is a fairly simple argument directly from the universal property. Given an abelian group $A$, define a group $F$ of "fractions" $\frac{a}{n}$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$. We impose an equivalence relation on these fractions by saying $\frac{a}{n}=\frac{b}{m}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(ma-nb)=0$. Using the usual formula for addition of fractions, you can check that the set $F$ of equivalence classes of fractions forms an abelian group.
There is now a bilinear map $f:A\times\mathbb{Q}\to F$ defined by $f(a,m/n)=\frac{am}{n}$. This induces a homomorphism $g:A\otimes\mathbb{Q}\to F$ which sends $a\otimes 1$ to the fraction $\frac{a}{1}$. So to show that $a\otimes 1\neq 0$, it suffices to show the fraction $\frac{a}{1}$ is nonzero. The zero element of $F$ is the fraction $\frac{0}{1}$, and $\frac{a}{1}=\frac{0}{1}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(1\cdot a-1\cdot 0)=0$, or in other words such that $ka=0$. So if $a$ is a non-torsion element, then $a\otimes 1\neq 0$. (In fact, this map $g$ is actually an isomorphism, but that takes more work to prove.)
The case of $\mathbb{R}$ follows from the case of $\mathbb{Q}$ since $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ and tensor products distribute over direct sums.
From a broader perspective, what's really going on here is that $\mathbb{Q}$ (or $\mathbb{R}$) is a flat $\mathbb{Z}$-module, meaning that if $A$ is a $\mathbb{Z}$-module and $B\subseteq A$ is a submodule, then the map $B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$ induced by the bilinear map sending $(b,q)\in\mathbb{Q}$ to $b\otimes q\in A\otimes\mathbb{Q}$ is injective. So, in particular, if $a\in A$ is any element such that $na\neq 0$ for all $n$, the submodule $B\subseteq A$ generated by $a$ is isomorphic to $\mathbb{Z}$. Since $\mathbb{Z}\otimes\mathbb{Q}\cong\mathbb{Q}$ and we have an injective map $\mathbb{Z}\otimes\mathbb{Q}\cong B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$, this implies $A\otimes\mathbb{Q}$ is nontrivial (and in fact $a\otimes1\neq 0$, since it is the image of the nonzero element $1\otimes 1\in \mathbb{Z}\otimes\mathbb{Q}$).
The argument using fractions above can be generalized to show that if $R$ is any commutative ring, then any localization of $R$ is flat as an $R$-module. For $\mathbb{Z}$-modules, flatness can be characterized quite simply in general: a $\mathbb{Z}$-module is flat iff it is torsion-free. See Show that a Z-module A is flat if and only if it is torsion free? for some sketches of the proof.
Best Answer
What about $A=\Bbb Q/\Bbb Z$? $A$ is both torsion and divisible, so $(1/m)\otimes (1/n)=n(1/nm)\otimes 1/n=(1/nm)\otimes (n/n)=0$.
ADDED IN EDIT
On the category of Abelian groups, Tor is left-exact in both arguments, so if $B$ is a subgroup of $A$ then $\text{Tor}(B,B)$ is a subgroup of $\text{Tor}(A,A)$. If $A$ has nontrivial torsion, then $A$ has a subgroup isomorphic to $C_n$, a cyclic group of finite order $n\ge 2$. Therefore $\text{Tor}(A,A)$ has a subgroup isomorphic to $\text{Tor}(C_n,C_n) \cong C_n$, and so is nonzero.