Answer
(i) Only one tension force $T_A$
(ii) Both $T_A$ and $T_B$, which oppose each other.
(iii) $v_A\ne v_B$
(iv) $T_A=15.6N$ and $T_B=23.4N$
Work Step by Step
(i) Only one tension force $T_A$ in the half-rod between A and B contributes to the centripetal force acting on A. In other words, $T_A=F_{cA}$
(ii) 2 tension forces opposing each other contribute to the centripetal force acting on B $F_{cB}$: $T_B$ supports $F_{cB}$, while $T_A$ reduces it. Therefore, $F_{cB}=T_B-T_A$
(iii) The speeds at different points with different radius $r$ are different from one another. So $v_A\ne v_B$
(iv) For $T_A$: $$T_A=F_{cA}=\frac{mv^2_A}{r_A}$$
We know $m=0.5kg$, $v_A=5m/s$ and $r_A=2L=0.8m$. Therefore, $$T_A=15.6N$$
For $T_B$: we do not know $v_B$ yet, so we need to find it.
First, we need period of rotation $T$, which is the same everywhere in this rotation: $$T=\frac{2\pi r_A}{v_A}=1.005s$$
As $r_B=L=0.4m$, we have $$v_B=\frac{2\pi r_B}{T}=2.5m/s$$
We know that $$F_{cB}=T_B-T_A$$ $$T_B=F_{cB}+T_A=\frac{mv^2_B}{r_B}+15.6N=23.4N$$
