Answer: A, B.
A)
Given: $(P_1 \land P_2 \land .....\land P_n) \land R \rightarrow S$ is a tautology which is equivalent to
${(P_1 \land P_2 \land .....\land P_n)'} \lor {R'} \lor S$ is a tautology.
Now the second part asks us if $(P_1 \land P_2 \land .....\land P_n) \models (R \rightarrow S) $ is valid.
If this argument is valid then $(P_1 \land P_2 \land .....\land P_n) \rightarrow (R \rightarrow S) $ is a tautology.
The above proposition is equivalent to $\underbrace{{(P_1 \land P_2 \land .....\land P_n)'}}_\text{LHS of first implication} \lor \underbrace{{R'}}_\text{LHS of second implication} \lor S$
We know from the given valid argument that this proprosition is a tautology, hence the second argument is valid.
B)
Let A: A wins, B: B places, C: C places, D: D places
Given argument:
$\begin{array}{ r l } & A \rightarrow {B \lor C} \\ & B \rightarrow {A'} \\ & D \rightarrow {C'} \\ & A \\ \hline \therefore & A \rightarrow {D'} \end{array}$
To check the validity of this argument we will try to make all the premises true assuming the conclusion is false.
$A \rightarrow {D'}$ can only be false when $A$ is true and ${D’}$ is false, ie. $D$ is true.
The 4th premise: $A$ is true.
To make the 3rd premise true $C$ will have to be false since $D$ is true.
To make the 2nd premise true $B$ will have to be false.
Now in the 1st premise $A$ is true while $B$ and $C$ both are false, so $B \lor C$ is false. Therefore this implication is false.
So we have failed to make all the premises true while the conclusion is false.
Thus the argument in question is valid.
Options C & D present the opposite case from option B & A respectively. Therefore, these options are incorrect.