It is not hard to check that $M(\mathbb M_n(\mathcal A))=\mathbb M_n(M(\mathcal A))$, and that $\mathbb M_n(\mathcal A)/\mathbb M_n(\mathcal J)=M_n(\mathcal A/\mathcal J)$ for any C$^*$-algebra $\mathcal A$ with ideal $\mathcal J$. Then $$ \mathbb M_n(M(\mathcal A)/\mathcal A)=\mathbb M_n(M(\mathcal A))/\mathbb M_n(\mathcal A)=M(\mathbb M_n(\mathcal A))/\mathbb M_n(\mathcal A). $$ That shows part 1. For part 2, we just apply part 1, and the isomorphism given: $$ \mathbb{M}_2(\cal{C}(\cal{A}\otimes\cal{K(H)}))=\mathcal C(\mathbb{M}_2(\cal{A}\otimes\cal{K(H)})\simeq \mathcal C(\cal{A}\otimes\cal{K(H)}) $$