in such type of questions set all 'premises' TRUE and then see if conclusion also becomes TRUE then given 'implication' is a valid formula.
a) $P$ $\vee$ $Q$ ------->>>premise 1
$ R\vee S$ ------->>>premise 2
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$(P\wedge R)\vee Q\vee S$ ---->>>Conclusion
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key concept is try to make conclusion false , and see if some of values for which our conclusion became FALSE , can make all premises TRUE?
we can observe to make conclusion $(P\wedge R)\vee Q\vee S$ FALSE , both $S$ and $R$ must have to be false and atleast one of $P$ and $R$ must be FALSE.so in that way our premises will become , $P\vee F and R\vee F$ which are nothing but $P$ and $R$ , so eventually formula will become $P\wedge R\Rightarrow P\wedge R$ , which is always TRUE , so this formula is a tautology.
b) $P\vee Q$ ------->>>Premise 1
$R\vee S$ -------->>>Premise 2
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$Q\vee S$ ------->>>Conclusion
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Make how can our conclusion be FALSE. It can be FALSE when both $Q$ and $S$ are FALSE. and in that way premises will become $P$ and $R$ , Now overall our formula reduced to $P\wedge R\Rightarrow F$. Here we can easily see , it can produce $T \Rightarrow F$ , when both $P$ and $R$ are TRUE.So this is clearly not a Tautology.
C) Here make $R\vee Q \vee S$ FALSE , which can only possible when all $R$,$Q$ and $S$ are FALSE and in that case our premises will become $P$ and FALSE , which will trivially make our formula tautology because $F\Rightarrow Anything$ is Tautology.
Same reasoning for other options.