1)
The accompanying simultaneous-move game is played twice, with the
outcome of the first stage observed before the second stage begins.
There is no discounting. The variable x is greater than 4, so that
(4,4) is not an equilibrium payoff in the one-shot game. For what
values of x is the following strategy (played by both players) a
subgame-perfect Nash equilibrium?

Play Qi in the first stage. If the first-stage outcome is (Q1,Q2),
play Pi in  the second stage. If the first-stage outcome is (y,Q2)
where y is not equal to Q1, play Ri in the second stage. If the
first-stage outcome is (Q1,z) where z is not equal to Q2, play Si in
the second stage. If the first-stage outcome is (y,z) where y is not
equal to Q1 and z is not equal to Q2, play Pi in the second stage.

      P2    Q2     R2    S2
P1   2,2   x,0   -1,0   0,0
Q2   0,x   4,4   -1,0   0,0
R3   0,0   0,0    0,2   0,0
S4   0,-1  0,-1  -1,-1  2,0

2)
Three oligopolists operate in a market with inverse demand given by
p(Q)=a-Q, where Q=q1+q2+q3 and qi is the quantity produced by firm i.
Each firm has a constant marginal cost of production, c, and no fixed
cost. The firms choose their quantities as follows:(1)firm 1 chooses
q1>=0; (2)firm 2 chooses 2 and 3 observe q1 and then simultaneously
choose q2 and q3, respectively. What is the subgame-perfect outcome?

(you might use the book "Game Theory for Applied Economists" which was
written by Robert Gibbons as a reference.)

First, I think you meant P1, Q1, R1, S1 in the first column. Correct
me if I am wrong, otherwise I am way off track.

As with any multistage game - start of by considering the smallest
subgame. In this case, that is the single stage game played at the
end.

Consider the optimal stragegy for player 1 given player 2's tactics :

If P2 then P1
If Q2 then P1
If R2 then R1
If S2 then S1

Likewise for player 2 given player 1 :

If P1 then P2
If Q1 then P2
If R1 then R2
If S1 then S2

By underlining these strategies on our payoff matrix (Can't do that
easily in text format) we see that the pure strategy nash equilibria
are :

(P1, P2) with payoff (2, 2)
(R1, R2) with payoff (0, 2)
(S1, S2) with payoff (2, 0)

So our three threats are "If you play ___ in the first round I will
play P/R/S"

We can basically now consider a two stage policy and find the values
of x for which neither party has an incentive to deviate.

As stated in the question, assume that playing Q is rewarded by
playing P in the second round, but that deviation from Q makes the
penalty route R or S be played. Then the updated payoff matrix for the
first round (contingent on equilibrium behavior in the second round)
is:

    P2      Q2      R2      S2
P1  4,4(P)  x,2(R)  1,2(P)  2,2(P)
Q1  2,x(S)  6,6(P)  1,0(S)  2,0(S)
R1  2,2(P)  0,2(R)  2,4(P)  2,2(P)
S1  2,1(P)  0,1(R)  1,1(P)  4,2(P)

(Second round stragegy given in parentheses).

Consider player 1's optimal choices given player 2 :
If P2 then play P1
If Q2 then play P1 if x>=6
           play Q1 if x<=6
If R2 then play R1
If S2 then play S1

Likewise for player 2 given player 1 :
If P1 then play P2
If Q1 then play P2 if x>=6
           play Q2 if x<=6
If R1 then play R2
If S1 then play S2

So if x<=6 then (P1P1, P2P2), (Q1Q1, Q2Q2), (R1P1, R2P2), (S1P1, S2P2)
are all equilibria.

However if x > 6 then (Q1Q1, Q2Q2) ceases to be an equilibrium.

So the answer to the question is that 4 < x <= 6.

Question 2 :

The oligopolists want to maximize (P(Q)-c)*qi.

Consider firm 2 and 3's decision based on q1 given :

Given that firm 1 has chosen q1 and firm 2 has chosen q2, firm 3
chooses q3 to maximize

q3*(a-q1-q2-q3-c)

ie. a-q1-q2-2q3-c = 0 (by taking derivative with respect to q3)

So q3 = (a-q1-q2-c)/2

Likewise, by symmetry q2 = (a-q1-q3-c)/2

Substituting q3 into the expression for q2 gives :

q2 = (a-q1-((a-q1-q2-c)/2)-c)/2

Rearranging we get :

q2 = (a-q1-c)/3

And by symmetry q3 = (a-q1-c)/3

So now we consider the decision of firm 1, given that q2,q3 =
(a-q1-c)/3

choose q1 to maximize :

q1(a - q1 - 2(a-q1-c)/3 - c)

or q1(a - q1 - c)/3

First order condition is : a/3 - 2q1/3 - c/3 = 0

ie. q1 = (a-c)/2

Substituting back into q2, q3 :

q2 = (a - (a-c)/2 - c)/3 = (a-c)/6

So the equilibrium outcome is :

q1 = (a-c)/2
q2 = (a-c)/6
q3 = (a-c)/6

Hopefully this all makes sense. Let me know if you need any part
clarifying

Regards

calebu2-ga

It is totally clear!
You are fabulous!
Thanks :)

Well, the book costs $35 for Amazon.

And Gibbons is a lot easier to read than Fudenberg and Tirole :)

I vowed never to study game theory again after taking a class in it. I
didn't realise my reservation price was so low :)

Calebu2-ga

Really?! Then i guess i am better to stick with this book. 
Both problems are actually from Gibbons, that's why.
Gees.....game theory makes me dizzy all the time.
Let me go over the problem and answer first to see if I get it or not. 
I'll be right back. :)
thanks ~

yup, i mean P1, Q1, R1, and S1.
sorry about the mistake.