These are three different theorems and there's no relation between 1) and the others except in the case when $X$ is finite, where all three theorems coincide (since $\ell^2(X)$, $C_0(X)$ and $C_c(X)$ then are the same topological vector space). Note however that $C_0(X)$ and $C_c(X)$ are never Hilbert spaces (unless the locally compact space $X$ is empty or a point), so 2) and 3) can't have a direct relation to 1).
Since positivity implies continuity 2) can be interpreted as characterizing continuous linear functionals on $C_c{(X)}$ as well, and I'm addressing this version below.
The results 2) and 3) are closely related and often 3) is proved as a corollary of 2).
Note that the space $C_c(X)$ is dense in $C_{0}(X)$ with respect to the sup-norm. So a continuous linear functional on $C_c(X)$ (= a signed Radon measure) extends (uniquely) to a continuous linear functional $C_{0}(X)$ (= a signed bounded Radon measure) if and only if it is of bounded variation. Moreover, 2) and 3) coincide if $X$ is compact (and there are also proofs of 3) reducing it to that case).
Well, for an arbitrary inner product on $X:=H^*$ it is not going to work, since then $X^*$ need not be isomorphic to $H$.
On the other hand, the Riesz representation gives a linear isomorphism $H\to H^*$, and if the inner product is defined via this isomorphism, i.e. if
$$(\langle x,-\rangle,\langle y,-\rangle)_{H^*}=\langle x,y\rangle_{H} $$
for all $x,y\in H, \ f\in H^*$, then your claim is valid:
Let $a,b\in H^*$ then $a=\langle x,-\rangle$ and $b=\langle y,-\rangle$ for some $y\in H$ by Riesz representation, and $a(y)=\langle x,y\rangle$ so we have
$$(a,b)_{H^*} = (a,\langle y,-\rangle)=a(y)=\langle a,y\rangle_{H^*,H}\ .$$
Best Answer
Here's a construction:
Take any reflexive Banach space $X$. Take the direct sum $E = X \oplus X^{\ast}$ and equip it with the norm $\|(x,x^{\ast})\|_E = \left(\|x\|_{X}^2 + \|x^{\ast}\|_{X^{\ast}}^2\right)^{1/2}$. This space is usually denoted by $E=X \oplus_2 X^{\ast}$ for short.
Then $E$ is isometric to its dual space $E^{\ast} = X^{\ast} \oplus_2 X^{\ast\ast}$: An isometric isomorphism is given by $(x,x^{\ast}) \mapsto (x^{\ast},i_{x})$ where $i:X \to X^{\ast\ast}$ is the canonical inclusion (by reflexivity of $X$ the map $i$ is an isometric isomorphism).
The space $E$ won't be isomorphic to a Hilbert space unless $X$ is itself isomorphic to a Hilbert space.
In fact, a (real) Banach space is “Hilbertable” if and only if every closed subspace has a complement by a (very deep) Theorem of Lindenstrauss and Tzafriri.
So: If $X$ is not isomorphic to a Hilbert space, it has a subspace that isn't complemented, and thus so has $E$.