This was initially intended to be a reply to geetha290krm's supposed counterexample to (2) (showing it is $\textbf{not}$ in fact a valid counterexample, and that in the context of $X$ being locally compact Hausdorff, (2) will always in fact be true), but was too long for a comment, so I wrote it up as an answer.
Lets work in $\mathbb{R}_{> 1}$, for simplicity (with the Lebesgue measure).
Let $0 \leq \chi_{n} \leq \frac{1}{n}$ be a bump function equal to $1/n$ on $[n + \epsilon_{n}, n+1 - \epsilon_{n}] $, with support contained in $[n + \epsilon_n^2 ,n + 1 - \epsilon_n^2] \subset (n,n+1)$ for each $n$ , where $\frac{1}{4}> \epsilon_{n} > 0$ are chosen sufficiently small so that $\epsilon_{n} \downarrow 0$ as $n \uparrow \infty$, finally notice that because $\frac{1}{4} > \epsilon_{n} > 0$, $(1 - 2 \epsilon_{n}) \geq \frac{1}{2}$ for all $n = 1,2,3,...$
Clearly then, $\chi_n \in C_c(\mathbb{R}_{>1}) \subset C_0(\mathbb{R}_{>1})$ for each $n$, and we have $|| \sum_{k=m}^n \chi_{k} ||_{u} \leq \frac{1}{m}$, for all $1 \leq m < n$, so that $\sum_{k \geq 1} \chi_{k} \rightarrow_{||.||_{u}} f$, where $f \in C_0(\mathbb{R}_{>1})$ and is non-negative everywhere.
Writing $\mathbb{R}_{>1} = (1,2) \cup (2,3) \cup (3,4) \cup ...$, we see that $\sum_{k \geq 1} \chi_k \restriction (n,n+1) = \chi_n$, as $\text{supp}(\chi_{k}) \subset (k,k+1)$ for all $k \geq 1$, so that only $\chi_n$ may be non-zero when $f$ is restricted to $(n,n+1) \subset \mathbb{R}_{>1}$.
geetha290krm claims that the measures $\mu_n = n d \lambda \restriction (n,n+1) $, which are indeed in $\mathcal{M}(\mathbb{R}_{>1})$ converge to $0$ in the weak$^{\ast}$-sense.
This is false, as we will now show. Indeed, suppose it were true. Then by the definition of convergence in the weak$^{\ast}$-sense as defined by the OP, we should have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} 0 = \int f d\mu$ , here $\mu = 0$ is the zero measure on $\mathbb{R}_{>1}$.
However, we have that $\int f d\mu_n \geq \int_{[n + \epsilon_n, n+1 - \epsilon_n]} f d\mu_n = (1 - 2 \epsilon_n)\frac{n}{n} \geq (1 - 2 \epsilon_n) \geq \frac{1}{2} $ for all $n = 1, 2 , 3 , ...$ with the inequalities all holding because $f$ is non-negative, its restriction to each interval $[n+\epsilon_n, n+1 - \epsilon_n] \subset (n,n+1)$ is just $\chi_n$ restricted to this interval, where it is identically equal to $\frac{1}{n}$, and because we chose $\frac{1}{4} > \epsilon_n > 0$ for all $n = 1,2,3,...$, we indeed have that $(1 - 2 \epsilon_n) \geq \frac{1}{2}$ for all $n = 1,2,3,...$ (as already established in the first paragraph).
This shows that we do not have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} \int f d \mu$, so that geetha290krm's counterexample is invalid, because $\mu_n$ does not converge to $0$ in the weak$^{\ast}$ - sense, as I have demonstrated.
Indeed, there is no counterexample to (2), by the Riesz-Markov theorem , whenever $X$ is a locally compact Hausdorff space, since then your $\mathcal{M}(X)$ is precisely isometric to $C_0(X)^{\ast}$, i.e. the dual space to $C_0(X)$, consisting of all bounded $\mathbb{R}$-linear functionals $C_0(X) \rightarrow \mathbb{R}$, which is a Banach space with the operator norm. By equipping $C_0(X)^{\ast}$ with the weak$^{\ast}$-topology, and then transporting this topology onto $\mathcal{M}(X)$ via the isometric isomorphism $\mathcal{M}(X) \rightarrow C_0(X)$ given by $\mu \rightarrow I_{\mu}$, where $I_{\mu}(f) = (f \rightarrow \int_{X} f d \mu)$, $I_{\mu} : C_0(X) \rightarrow \mathbb{R}$, your question for (2) is answered in the affirmative, because it is a general fact that for a Banach space $X$, a weak$^{\ast}$-convergent sequence $f_n \in X^{\ast}$ is uniformly bounded in operator norm.
Therefore to seek a counterexample for (2), we need to find a (metrizable) Hausdorff topological space $X$, which is NOT locally compact.
One candidate is $X = (1,2) \times (2,3) \times (3,4) \times (4,5) \times ...$, with the usual product metric (inducing the product topology). This is NOT locally compact, so perhaps we could find a counterexample here.
Now perhaps it is possible to adapt geetha290krm's counterexample to this situation. But I have not figured out all the details yet and am not entirely convinced $X$ is sufficient to provide a counterexample to (2) for the following reason:
While this space $X$ is not locally compact, it is Borel isomorphic to $[0,1]$, which is indeed (locally) compact Hausdorff, so perhaps this $X$ may not work to produce a counterexample.
So for (2) a counterexample may need to come from a metrizable topological space $Y$, that is NOT locally compact, and is either
1: NOT second countable, or
2: NOT $\textbf{completely}$ metrizable if it $\textbf{is}$ second countable (i.e. if condition 1. does hold)
Either of these conditions will prohibit $Y$ from being Borel isomorphic to a Borel subset of $[0,1]$, where (in the latter space $[0,1]$) the usual Riesz-Markov theorem applies, and this may cause complications.
Best Answer
Firstly, we remark that $\mathcal{M}(X)$, as a set, is the collection of all finite signed Radon measures (assuming that we are working with the scalar field $\mathbb{R}$). That is, $\mu\in\mathcal{M}(X)$ if $\mu$ is a finite signed measure, outer-regular in the sense that for each Borel subset $A\subseteq X$, $\mu(A)=\inf\{\mu(U)\mid A\subseteq U\mbox{ and }U\mbox{ is open}\}$, and inner-regular for open set in the sense that for each open set $U\subseteq X$, $\mu(U)=\sup\{\mu(K)\mid K\subseteq U\mbox{ and }K\mbox{ is compact.}\}$. Note that if $X$ is second countable, then every finite signed measure is automatically Radon. (see Folland, Real Analysis, Theorem 7.8).
For your question, the two modes of convergence are not the same. Obviously, given a sequence $(\mu_{n})$ in $\mathcal{M}(X)$ and $\mu\in\mathcal{M}(X)$, $\mu_{n}\rightharpoonup\mu\Rightarrow\mu_{n}\rightarrow\mu$ in weak*-topology. The converse does not hold in general. The following is a counter-example.
Let $X=\mathbb{R}$, equipped with the usual topology. For each $n\in\mathbb{N}$, let $\mu_{n}(A)=\lambda(A\cap[n,n+1])$, where $\lambda$ is the usual Lebesgue measure, then $\mu_{n}\in\mathcal{M}(X)$. Let $\mu=0$. Clearly $\mu_{n}\rightarrow\mu$ in weak*-topology. However, $\mu_{n}\not\rightharpoonup\mu$ because the constant function $f=1$ is an element in $C_{b}(X)$ and $\int fd\mu_{n}=1\neq\int fd\mu$.