For these questions it is better to get some counter examples by trying small graphs. Consider the following graph:
Now, as per the question, an arborescence is a subgraph with all incoming edges to the special vertex $r.$ So, for the above graph it is just the edge $a-r$ which is of weight $2.$
Now, option A says that if we sum up outgoing the edge weights (minimum weight in case multiple edges exist) of all vertices excluding $r$, the sum will be greater than or equal to the weight of the minimum arborescence. This is true because all the edges in the minimum arborescence will be including as the outgoing edges of the other vertices.
Now, option B is false. Because instead of outgoing edges we are adding the weights of incoming edges. And these edges are not part of the arborescence. For example, in the above graph, only one vertex $a \in V \backslash \{r\}.$
$ \sum_{u \in V \backslash \{r\}} \min_{(v,u) \in E} w((v,u)) = w(r,a) = 5 > w^*.$
Options C and D are straightforward TRUE as for minimum arborsecence we only need to consider the incoming edges to $r$ from $n-1$ other vertices thus making it acyclic too.
For option E, suppose a graph has a directed Hamiltonian cycle and its weight is $\leq$ $w^*.$ Since Hamiltonian cycle on a graph of $n$ vertices must consist of $n$ edges and $H^*$ has only $n-1$ edges and all edge weights being positive, this means there is at least one edge $e'$ in $H^*$ which is having a weight larger than any other edge in Hamiltonian cycle. Is this possible? Consider the below graph:
Here, Hamiltonian cycle is $a - b - c - r - a$ and its weight will be $1+1+2+1 = 5.$
Minimum arborescence (shown by thick edges) has the edges $a-r,b-r,c-r$ and its weight will be $2+100+2 = 104.$ So, we have a counter example for option E.
So, Correct answer: B, E.