Partial answer: I will give a proof assuming that $E$ is separable. Of course this will give a proof when $E$ is not separable but $\mu$ has separable support.
It is an interesting fact that if support of $\mu$ exists in the sense that there is a smallest closed set of full measure then it is necessarily separable. [This requires Axiom of Choice]
Under this hypothesis it is known that $\mu$ is tight. Ref. Convergence of Probability Measures by Billingsley.
Let $\epsilon >0$ and choose a compact set $K$ such that $\mu (K^{c}) <\epsilon$. Then $$|\phi (x')-\phi (y')|$$ $$ \leq \int |e^{i \langle x', x \rangle}-e^{i \langle x', x \rangle}| d\mu (x)$$ $$\leq \int_K |e^{i \langle x', x \rangle}-e^{i \langle x', x \rangle}| d\mu (x)+2\epsilon.$$ So $$|\phi (x')-\phi (y')| \leq \|x'-y'\|\int_K \|x|| d\mu(x)+2\epsilon<3\epsilon$$ if $$\|x'-y'\| <\frac {\epsilon} {M\mu(E)}$$ where $$M=\sup \{\|x\|:x \in K\}$$.
Q1: Yes, $E_1'$ is a locally convex space when endowed with the $\sigma(E_1',E_1)$-topology. (It is just $E_2$ with the $\sigma(E_2,E_1)$-topology.)
Q2: The weak-$*$ topology respects subspaces. In general, if $\langle E, G \rangle$ is a dual pair and if $F \subseteq G$ is a subspace, then the relative $\sigma(G,E)$-topology on $F$ coincides with the $\sigma(F,E/F^\perp)$-topology.¹ If $F$ separates points on $E$, so that $\langle E,F\rangle$ is again a dual pair, then one has $F^\perp = \{0\}$, so now the $\sigma(F,E/F^\perp)$-topology is simply the $\sigma(F,E)$-topology.
¹: For a textbook reference, see [Sch99, §IV.4.1, Corollary 1 (p. 135)] or [Köt83, Proposition 22.2.(1) (p. 276)], among others. This is basically just a special case of the transitive law of initial topologies (see halfway through this very long answer), since weak topologies and subspace topologies are examples of initial topologies.
In your example, we would have $E = C_b(\Omega)$, $F = \mathcal M(\Omega)$ and $G = C_b(\Omega)'$. It is easy to see that $\mathcal M(\Omega)$ separates points on $C_b(\Omega)$, so $F^\perp = \{0\}$. Furthermore, we know from this other question of yours that $C_b(\Omega)$ separates points on $\mathcal M(\Omega)$, so $\mathcal M(\Omega)$ can be viewed as a subspace of $C_b(\Omega)'$. Hence we are in the setting described above.
References
[Köt83] Gottfried Köthe, Topological Vector Spaces I, Second revised printing, Grundlehren der mathematischen Wissenschaften 159, Springer, 1983.
[Sch99] H.H. Schaefer, with M.P. Wolff, Topological Vector Spaces, Second Edition, Graduate Texts in Mathematics 3, Springer, 1999.
Best Answer
To put this result into context, I will show how to deduce it from the big theorems. (This is not nearly as elementary as Kavi's proof.)
Unfortunately, different authors use different notation / terminology, so let me set up a few definitions first.
If $\Omega$ is a topological space, let $\mathcal A_\Omega$ and $\mathcal B_\Omega$ denote respectively the algebra and the $\sigma$-algebra generated by the open (or closed) sets of $\Omega$. Then $\mathcal A_\Omega \subseteq \mathcal B_\Omega$, and $\mathcal B_\Omega$ is the $\sigma$-algebra generated by $\mathcal A_\Omega$, so every measure on $B_\Omega$ is uniquely determined by its values on $A_\Omega$.
We will use the following well-known results:
It follows that the space $\mathcal M(\Omega)$ of finite signed Borel measures is a subspace of $rba(\Omega) \cong C_b(\Omega)'$. To complete the proof, note that every normed space $X$ separates points on every subspace of $X'$: if $\varphi(x) = 0$ for all $x \in X$, then $\varphi = 0$.
(More generally, for a bilinear pairing $\langle E , F \rangle$ to be non-degenerate, so that it is a proper dual pairing, it is necessary and sufficient that the induced maps $E \to F^*$ and $F \to E^*$ are injective.)
References.
[DS58] Nelson Dunford, Jacob T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.
[AB06] Charalambos D. Aliprantis, Kim C. Border, Infinite Dimensional Analysis, A Hitchhiker's Guide, Third Edition, Springer, 2006.