[7.] Not sure what you mean here. Are you talking about Noetherian groups? If so, then you can show that any subgroup $H$ of $G$ is finitely generated, given that $K$ and $G/K$ are Noetherian for some $K\trianglelefteq G$. This follows from the facts that $H\cap K\leq K$ is finitely generated and that $H/(H\cap K)\cong HK/K\leq G/K$ is also finitely generated.
[8.] Not sure what you mean here either. Are you talking about Artinian groups?
If so, the answer is also yes. Consider the chain of normal subgroups $G=G_0\geq G_1\geq G_2\geq \ldots$. Suppose that a normal subgroup $K$ of $G$ satisfies $K$ and $G/K$ being Artinian. The image of this chain under $G\to G/K$ is given by $G/K=H_0\geq H_1\geq H_2\geq \ldots$. Then, by the assumption that $G/K$ is Artinian, we get $H_q=H_{q+1}=H_{q+2}=\ldots=:\bar{H}$ for some $q\in\mathbb{N}_0$. Now, we look at the sequence of subgroups of $K$, $K=K_0\geq K_1\geq K_2\geq \ldots$, where $K_i:=K\cap G_i$ for $i=0,1,2,\ldots$. Note that, for some integers $k\geq q$, $K_k=K_{k+1}=K_{k+2}=\ldots=:\bar{K}$. Thus, for $l=k,k+1,k+2,\ldots$, we have the exact squence $1 \to \bar{K} \to G_l \to \bar{H}\to 1$. By the Five Lemma for groups, $G_k=G_{k+1}=G_{k+2}=\ldots$.
[9.] If $1\to K \to G \to Q\to 1$ is exact with $K$ and $Q$ being periodic, then every $x\in G$ satisfies $x^q\mapsto 1_Q$ via $G\to Q$ for some integer $q>0$. Thus, $x^q$ is in the image of $K\to G$. As $K$ is periodic, $x^{qk}=\left(x^q\right)^k=1_G$ for some integer $k>0$.
[10.] If $1\to K\to G\to Q\to 1$ is exact with $K$ and $Q$ being torsion-free, then $x\in G$ with $x^q=1_G$ for some integer $q>0$ implies $x^q\mapsto 1_Q$ via $G\to Q$. As $Q$ is torsion-free, $x\mapsto 1_Q$ via $G\to Q$. Hence, $x$ is in the image of $K\to G$. Since $K$ is torsion-free and the map $K\to G$ is injective, $x^q=1_G$ happens only when $x=1_G$.
