I am reading the following article
Jean-Pierre Gossez. “On the range of a coercive maximal monotone operator in a nonreflexive Banach space”. In: Proc. Amer. Math. Soc. 35 (1972), pp. 88–92
in which the following linear operator $A:\ell_1(\mathbb{N})\to\ell_\infty(\mathbb{N})$ is defined by
$$ (Ax)_n:=-\sum_{k<n}x_{k}+\sum_{k>n}x_{k}=\sum_{k\in\mathbb{N}}x_{k}\alpha(k,n),\ x=(x_{k})_{k\in\mathbb{N}}\in\ell_{1},\ n\in\mathbb{N}, $$
where $\alpha(k,n)=\left\{ \begin{array}{lll} -1, & \mathrm{if} & k<n\\ 0, & \mathrm{if} & k=n\\ 1, & \mathrm{if} & k>n \end{array}\right.$ , $k,n\in\mathbb{N}$.
Also, the adjoint of $A$ is considered $A^*:\ell_{\infty}^{*}\to\ell_{\infty}$ which is given by $\langle y,A^{*}\mu\rangle=\langle\mu,Ay\rangle$, for $y\in\ell_{1}$, $\mu\in\ell_{\infty}^{*}$, where $\langle\cdot,\cdot\rangle$ is the natural duality formed by a Banach space $X$ with its topological dual $X^*$, $\langle x,x^*\rangle:=x^*(x)$.
The author says "It is easy to verify that" $$ \langle\mu,-A^{*}\mu\rangle=(\mu(\beta\mathbb{N}\smallsetminus\mathbb{N}))^{2}.\hspace{3cm}(\star) $$ Here $\beta\mathbb{N}$ is the Stone-Cech compactification of $\mathbb{N}$, $\ell_{\infty}^{*}$ is identified with $M(\beta\mathbb{N})$ the space of finite Borel measures on $\beta\mathbb{N}$, and $\mu(\beta\mathbb{N}\smallsetminus\mathbb{N})$ denotes the measure of the set $\beta\mathbb{N}\smallsetminus\mathbb{N}$. I am mainly interested in the proof of ($\star$).
I tried getting ($\star$) by finding the expression of $A^*$ as follows. Every $x\in\ell_\infty$ is uniquely extended to (identified with) $\beta x\in C(\beta\mathbb{N})$ while $C(\beta\mathbb{N})^*$ is identified with $\ell^*_\infty$ via Riesz's Representation Theorem as follows: every $\mu\in C(\beta\mathbb{N})^*$ is a finite Borel measure (on $\beta\mathbb{N}$) such that $\mu(f)=\int f d\mu$, for $f\in C(\beta\mathbb{N})$.
I wonder whether for $\mu\in \ell^*_\infty$, $x\in\ell_\infty$, $\mu(x)=\sum_{n\in\mathbb{N}}x_n \mu(\{n\})$, where $\mu(\{n\}$ denoted the measure of the set $\{n\}$?
Any other way to approach ($\star$)?
As for your second question
The answer is no, not every measure on $\beta \mathbb N$ is discrete. For example, take a continuous surjection $f\colon \beta \mathbb N\to [0,1]$. Then $\mu(A) = m(f[A])$ ($A$ is a Borel subset of $\beta \mathbb N$), where $m$ is the Lebesgue measure, is an atomless measure so it cannot have the form you specified.
Compact spaces on which every measure is discrete (in particular, of the form you specified) are precisely the scattered spaces. (This was noted by W. Rudin in the 1950s.)
Every measure $\mu$ however naturally decomposes into
$$\langle \mu ,f \rangle = \sum_{n=1}^\infty f_n \mu(\{n\}) + \int_{\beta N\setminus N} (\beta f)\,{\rm} d\mu\quad (f\in \ell_\infty)$$
so the first step is to show that the former summand always vanishes for $\langle \mu, -A^*\mu\rangle$.