I am trying to read the book C$^*$-algebras and finite dimensional approximations.
Definition 2.6.1. A group $\Gamma$ is amenable if there exists a state $\mu$ on $\ell^{\infty}(\Gamma)$ which is invariant under the left translation action : for all $s\in\Gamma$ and $f\in \ell^{\infty}(\Gamma)$, $\mu (s\cdot f)=\mu (f)$.
Definition 2.6.2. For a discrete group $\Gamma$. we let $Prob(\Gamma)$ be the space of all probability measures on $\Gamma$ : $$\text{Prob}(\Gamma)= \text{Prob}(\Gamma)=\{\mu\in \ell^{1}(\Gamma):\mu\geq {0}\ \text{ and }\sum_{t\in\Gamma} \mu(t)=1 \}.$$
Note that the left translation action of $\Gamma$ on $\ell^{\infty}(\Gamma)$ leave the subspace $\text{Prob}(\Gamma)$ invariant , hence we can also use $\mu \to s\cdot\mu $ to denote the canonical action of $\Gamma$ on $\text{Prob}(\Gamma)$.
Definition 2.6.1 $\implies$ Definition 2.6.2.
Proof: Take an invariant mean $\mu$ on $\ell^{\infty}(\Gamma)$ .Being the predual of $\ell^{\infty}(\Gamma)$, $\ell^{1}(\Gamma)$ is dense in $\ell^{\infty}(\Gamma)^*$ and thus we can find a net $\mu_{i}$ in $\text{Prob}(\Gamma)$ which converges to $\mu$ in the Weak star topology . note that for each $s\in\Gamma$, the net $(s.\mu_{i}-\mu_{i})$ converges to zero weakly in $\ell^{1}(\Gamma)$. Hence, for any finite subset $E\subset \Gamma$, the weak closure of the convex subset $\bigoplus_{s\in E}\{s.\mu-\mu:\mu\in \text{Prob}(\Gamma)\}$ contains zero. Since the weak and norm closures coincide, Definition 2.6.2 follows.
Question (1) is how to understand the net $(s.\mu_{i}-\mu_{i})$ converges to zero weakly in $\ell^{1}(\Gamma)$. we know that $\mu_{i}$ in $Prob(\Gamma)$ which converges to $\mu$ in the Weak star topology (i.e $f(\mu_{i}) \to f(\mu)=\mu(f)$, for every $f\in \ell^{\infty}(\Gamma)\subset \ell^{\infty}(\Gamma)^{**}$ ) ).Since $\mu(f) = \mu(s.f) $ (by def 2.6.1) , but i don,t understand the relation $\mu(s.f)$ with $f(s.\mu_{i})$. if $ s.f(\mu_{i})=f(s.\mu_{i})\to \mu(s.f)=\mu(f) $, then we done. but i don,t known how to show $ s.f(\mu_{i})=f(s.\mu_{i})$ . please help me !!!
Question (2) : how to prove $(\ell^{\infty}(\mathbb I_{1})\bigoplus \cdots\bigoplus\mathbb{\ell^{\infty}(I_{k})})^{**}\cong \ell^{\infty}(\mathbb I_{1})^{**}\bigoplus \cdots\bigoplus\mathbb{\ell^{\infty}(I_{k})}^{**}$.
You have that $\mu_j\to\mu$ in the weak$^*$-topology. This means that $\mu_j(f)\to\mu(f)$ for all $f\in\ell^\infty(\Gamma)$. The duality here is given by $$ \mu_j(f)=\sum_{t\in\Gamma}f(t)\mu_j(t). $$ We have that $s\cdot\mu_j-\mu_j$ is in $\ell^1(\Gamma)$, so the weak$^*$-topology (as elements of $(\ell^1(\Gamma))^{**}$) is given by evaluation on $f\in\ell^\infty(\Gamma)$, which is precisely the weak topology.
Since $\mu_j\to\mu$, given $s\in\Gamma$, $f\in\ell^\infty$ and $\varepsilon>0$ there exists $j_0$ such that $|\mu_j(f)-\mu(f)|<\varepsilon/2$ and $|\mu_j(s^{-1}\cdot f)-\mu(s^{-1}\cdot f)|<\varepsilon/2.$
So, for $j\geq j_0$, \begin{align} |(s\cdot\mu_j-\mu_j)(f)|&=\left|\sum_t f(t)\mu_j(st)-f(t)\mu_j(t)\right|\\ \ \\ &\leq\left|\sum_t f(t)\mu_j(st)-\mu(s^{-1}\cdot f)\right|+\left|\mu(s^{-1}\cdot f)-\sum_t f(t)\mu_j(t)\right|\\ \ \\ &=\left|\sum_t f(s^{-1}t)\mu_j(t)-\mu(s^{-1}\cdot f)\right|+\left|\mu(f)-\sum_t f(t)\mu_j(t)\right|\\ \ \\ &=|\mu_j(s^{-1}\cdot f)-\mu(s^{-1}\cdot f)|+|\mu(f)-\mu_j(f)|\\ \ \\ &<\frac\varepsilon2+\frac\varepsilon2=\varepsilon. \end{align}
As for your second question, it is a general fact that given two normed spaces $V,W$, one has $(V\oplus W)^*\simeq V^*\oplus W^*$.
Indeed, define $\gamma:(V\oplus W)^*\to V^*\oplus W^*$ in the following way. Given $f\in (V\oplus W)^*$, let $f_V\in V^*$ be given by $f_V(v)=f(v\oplus 0)$. Similarly, $f_W(w)=f(0\oplus w)$ defines an element of $W^*$. So we put $\gamma(f)=f_V\oplus f_W$. It is clear that $\gamma$ is linear. It is also bounded: using the norm $\|f\oplus g\|=\max\{\|f\|,\|g\|\}$, we have $$ \|\gamma(f)\|=\|f_V\oplus f_W\|=\max\{\|f_V\|,\|f_W\|\}\leq\|f\|. $$ The last inequality above comes from $$|f_v(v)|=|f(v\oplus 0)|\leq\|f\|\,\|v\oplus0\|=\|f\|\,\|v\|. $$ If $\gamma(f)=0$, then $f_V=0$ and $f_W=0$. Then $$f(v\oplus w)=f(v\oplus0)+f(0\oplus w)=f_V(v)+f_W(w)=0.$$So $\gamma$ is one-to-one. Finally, given $g\in V^*$ and $h\in W^*$, define $f(v\oplus w)=g(v)+h(w)$. Then $\gamma(f)=g\oplus h$. So $\gamma$ is onto.