Jacobson radical of group rings 
$4 b)$
(i) Any hints? 
(ii) Well $R$ is not semi-simple since $|\mathbb{Z}/3|=3=0 \in F_3$ by the converse of Maschke's theorem.
(iii) The surjective $\mathbb{C}$-algebra map $\phi:R \to M_2(\mathbb{C})\times\mathbb{C}\times\mathbb{C}: (a_{i,j}) \mapsto \begin{bmatrix}
    a_{11}       & a_{12} \\
    a_{21}       & a_{22} 
\end{bmatrix},a_{33},a_{44}$ has nilpotent kernel and semi-simple target.Hence the kernel is the Jacobson radical i.e. \begin{bmatrix}
    0       & 0 & *  & * \\
    0      & 0 & * & * \\
    0      & 0 & 0 & 0 \\
    0       & 0 & 0  & 0
\end{bmatrix}
Sample solution that may help with iii)


(i) If you know Maschke's theorem (as you hint in (ii)) then you already know the answer to this since the order of $D_8$ is $8$.
(ii) Yes. And it would not even be too hard to exhibit some nonzero nilpotent element to prove that the Jacobson radical is nonzero.
(iii) You can use exactly the same logic here at this similar question.  I'm not sure about your attempt. A surjective homomorphism from $R\to M_2(\mathbb C)\times \mathbb C\times \mathbb C$ would have a six dimensional image and a $6$ dimensional kernel (not $4$ dimensional).  But part of the logic at the other post confirms that you can find a homomorphism onto $M_2(\mathbb C)\times M_2(\mathbb C)$, which has and $8$-dimensional image and the $4$-dimensional kernel you describe. Since the quotient has Jacobson radical zero, that is excactly the Jacobson radical of $R$, then.
