Let $A$ and $B$ simple $C^*$-algebras. One can prove that the minimal tensor product $A\otimes _{min}B$ is simple. This is wrong for the maximal tensor product $A\otimes_{max}B$ .

1.Do you know an explicit example?

One idea is to consider (non-nuclear) $C^*$-algebras $A$ and $B$, which are simple and then consider the canonical map $i:A\otimes_{max} B\to A\otimes_{min}B$. One can take $A$ and $B$ such that the map has non-trivial kernel to obtain $\ker(i)\subseteq A\otimes_{max} B$ as a non-trivial closed ideal. Also the example should satisfy: $\ker(i)\neq A\otimes_{max} B$.

Therefore if you take this idea, for example, $A=K(H)=B$ for a separable Hilbert space $H$ doesn't work because these $C^*$-algebras are nuclear, simple and the minimal tensor product is simple.

Then I tried to take (non-nuclear) group $C^*$-algebras. But I'm not sure for which groups $\Gamma$ the reduced group $C^*$-algebra $C_r^*(\Gamma)$ is non-simple, my knowledge is still very limited. For example, there is a result in a paper that if $\Gamma$ is a nonabelian free group, the $C_r^*(\Gamma)$ is simple (R.T. Powers, "Simplicity of the $C^*$-algebra associated with the free group on two generators.").

Therefore, my 2. question is: Do you know a (non-abelian) group $\Gamma$, such that $C_r^*(\Gamma)$ is not amenable and non-simple?