Idempotent of closure and other properties
It would be helpful if you provided the relevant definitions, but maybe what follows will be helpful.
It is generally true that if $F$ is closed then $\mathrm{Cl} F = F$. If you define $\mathrm{Cl} A$ as the smallest closed set containing $A$, then this is obvious: $\mathrm{Cl} F$ has to contain $F$, but already $F$ is closed so this is the smallest thing possible. If you have a different definition, try proving it's equivalent to the one above.
How is $\mathrm{diam} A$ defined? It is a supremum of some distances. Surely, a larger set has a larger diameter, so you just need one inequality. Suppose that $x_n,y_n \in \mathrm{Cl}A$ are such that $d(x_n, y_n) \to \mathrm{diam} \mathrm{Cl}A$. Try choosing $x'_n$ and $y'_n$ in $A$, close to $x_n,y_n$ (say, $d(x_n,x'_n) < 1/n$. What happens then?
Are you sure it's true? Try $X = \mathbb R$, $G = \{x \ : \ x < 0 \}$ and $A = \{x \ : \ x > 0 \}$ .
I assume that $(X,d)$ is a metric space and that $\text{Cl}A$ is defined as the set of all adherence points. If $x\in\text{Cl}\text{Cl}A$, then each open neighborhood $U$ of $x$ intersects $\text{Cl}A$, so there is a $y\in U\cap \text{Cl}A$. But $U$ is also a neighborhood of $y$, so there is a $z\in U\cap A$. This shows that $U$ intersects $A$, so $x$ is also an adherence point of $A$.
The diameter $\text{diam}A$ is defined as $\sup\{d(x,y)\mid x,y\in A\}$. Assume that $\text{diam}\text{Cl}A=D>d=\text{diam}A$. Choose $0<\epsilon<(D-d)/3$. There is a pair $x,y\in\text{Cl} A$ such that $d(x,y)>D-\epsilon$. As $x,y$ are in the closure of $A$, there are points $x',y'\in A$ such that $d(x,x')$ and $d(y,y')$ are less that $\epsilon$. Now
$$d(x',y') \ge d(x,y)-d(x,x')-d(y,y')\\ >(D-\epsilon)-\epsilon-\epsilon =D-3\epsilon\\ >D-(D-d)=d$$
For (3): If $x\in\text{Cl}(G\cap\text{Cl}A)$, then an open neighborhood $U$ of $x$ intersects $G\cap\text{Cl}A$. Then $U\cap G$ is open and $(U\cap G)\cap\text{Cl}A$ is non-empty, so $(U\cap G)\cap A=U\cap(G\cap A)$ is non-empty, which shows that $x\in\text{Cl}(G\cap A)$