Logical structure of arguments. So here are the contextual statements:
1) Maya either listens to music or does her homework. If she listens to music she feels happy.If she does her homework she feels unhappy. Therefore she will not do her homework while listening to music.
Let P be the statement "Maya listens to Music".
Q "Maya does homework".
R "Maya feels happy".
So am I right to write it as ((P=>Q)∨(Q=>¬R) ) => ( ¬Q∧¬P ) 
2) If I drink coffee, then I will get my assignment done on time. If I do not drink coffee, then I will feel sleepy. If I feel sleepy, then I will make mistakes. Therefore, if I will not get the assignment done on time, then I will make mistakes.
Let P be the statement "I drink coffee".
Q "I get the assignment done on time".
R "I will feel sleepy".
S "I will make mistakes".
( (P=>Q)∧(¬P=>R)∧(R=>S) ) =>( ¬Q=>S ) . Is it right?
In order to examine whether the arguments are right do I really need to do the truth tables..? It will be a huge one for the second statement.
 A: In (1) you're missing a translation of "Maya either listens to music or does her homework". Otherwise they look right at first glance [edit: see the other answer, though; there was a mistake I missed]. But you could help readability a lot by choosing variable letters that have something to do with their meaning (say, the first letter of a key word in the statement) instead of just picking $P$, $Q$, $R$, $S$ in sequence.

In order to examine whether the arguments are right do I really need to do the truth tables..?

No, you don't in general need to do proof table. That's the slow and cumbersome, but reliable method that you can always use and which doesn't need any particular smarts.
However, it often happens that you can find some kind of shorter argument that works as well (unless, as may be the case, you're specifically asked to use truth tables for educational reasons).
If the argument is not valid, then it's enough to show a single counterexample -- you don't have to find it systematically by filling out a truth table; any kind of inspired guesswork is allowed as long as what it ends up with is in fact a counterexample.
If the argument is valid, then it will be possible to give a proof that it is valid. At the level where one is typically asked to solve exercises like these, you won't yet have learned any system for writing and recognizing formal proofs -- but part of the goal of a course that contains this kind of exercise will be to teach you to recognize the shape of the informal kind of arguments that actual mathematics uses as proof, and get enough familiarity with them that you can recognize a valid argument without drawing up a truth table.
A: 1)
$\underbrace {(P \rightarrow R)}_{\text{Listening to music makes Maya happy}} \land \underbrace {(Q \rightarrow \lnot R)}_{\text{Doing Homework Makes Maya Unhappy}} \rightarrow \underbrace {\lnot (P \land Q)}_{\text{Maya won't be doing homework and listening to music}}$
2)
$(\underbrace{P \rightarrow Q}_{\text{Coffee then on time}} \land \underbrace{\lnot P \rightarrow R}_{\text{No coffee then sleepy}} \land \underbrace{R \rightarrow S}_{\text{sleepy then mistakes}}) \rightarrow (\underbrace { \lnot  Q \rightarrow S }_{\text{Not on time then mistakes}})$
Assume the above statement is false.  The only false example of an implication is $true \rightarrow false$.  Therefore:
$(P \rightarrow Q \land \lnot P \rightarrow R \land R \rightarrow S) \text { is true and} \tag{1}$
$ \lnot  Q \rightarrow S \text{ is false} \tag{2}$
From (2) you get $Q$ is false and $S$ is false.  (1) becomes
$P \rightarrow Q \tag{3}$
$\lnot P \rightarrow R \tag{4}$
$R \rightarrow S \tag {5}$
From (3) and false $Q$ you get that $P$ is false.
From (5) and false $S$ you get that $R$ is false.
Then (4) becomes $true \rightarrow false$, which is the contradiction that proves that the original statement is true.
