Note: there is more than one right answer to many of these questions.
You may receive full marks even if your solution looks very different.

1. We prove that L(G1) U L(G2) is context-free by exhibiting a CFG
for it, called <Sigma,N,P,S>.  Without loss of generality, we may
assume that N1 and N2 are distinct, i.e., have no intersection.
If not, we may consistently rename the non-terminals of one of the 
grammars to establish this.  Let Sigma = Simga1 U Sigma2, let 
N = N1 U N2 U {S}, where S is a new symbol (not in N1 nor N2).  Let 
P = P1 U P2 U {S -> S1 | S2}.  A string w belongs to L(G1) U L(G2) iff
S1 =>* w under G1 or S2 =>* w under G2.  The latter are derivable 
under G from S because S => S1 =>* w and/or S => S2 =>* w.

2. G1: <{a,b},{S,X,Y},{S -> X Y, X -> a b | a X b, Y -> c | c Y},S>
   G2: <{a,b},{S,X,Y},{S -> X Y, X -> a | a X, Y -> b c | b Y c},S>

L(G1) = {a^n b^n c+ | n >= 1}, L(G2) = {a+ b^n c^n | n >= 1}.  So
L(G1) ^ L(G2) = {a^n b^n c^n | n >=1}, which is not context-free.

3. (a) There are two:
(1)
S
|
|- S - 3
|- Op - +
 - S
   |- S - 1
   |- Op - x
    - S - 2
(2)
S
|
|- S
|  |- S - 3
|  |- Op - +
|  |- S - 1
|- Op - x
|- S - 2

(b) Yes. It assigned the two trees in (a) to 3+1x2.
(c) No. See (d).  The grammar presented there assigns one parse
to every string in L(G), and assigns parses to only the strings of
L(G).
(d) <{0,1,2,3,+,-,x,/,(,)},{S, Op},P,S>, where
P = {S -> Term TermOp S | Term,
     Term -> Factor FactorOp Term | Factor,
     Factor -> Num | ( S ),
     Num -> 0 | 1 | 2 | 3,
     TermOp -> + | -,
     FactorOp -> x | /}.

4. G = <Sigma,{1,...,n},P,1>, where
P = { s0_i -> tji s1_i | <s0_i,tji,s1_i> in E }
  U { s0_i -> tji | <s0_i,tji,s1_i> in E and s1_i in F }
  U { epsilon | if 1 in F}

5. <{a,b,c,d},{S,X,Y},P,S>, where
P = {S -> X Y,
     X -> a b | a X b,
     Y -> c d | c Y d}

6. <{a,b,c,d},{S,X},P,S>, where
P = {S -> a X d | a S d,
     X -> b c | b X c}

7. <{a,b,c,d},{S},P,S>, where
P = {S -> EvenNM,
     EvenNM -> a OddNM d | a OddM d,
     OddNM -> a EvenNM d | a EvenM d,
     OddM -> b c | b EvenM c,
     EvenM -> b OddM c}

8. {a^i b^n c^j | i,j >= 1, i+j=n} = {a^i b^i b^j c^j | i,j >= 1}, so
the answer to this is the same as for Question 5, substituting a -> a,
b -> b, c -> b, d -> c.

9. <{a,b},{S},{S -> epsilon | a S b | b S a | S S},S>
[a modification of this that does not generate the empty string is also 
 acceptable]

10.  "The intersection of a CFL and a regular language is always
context-free."  This is the fact that we are allowed to assume.  By
contraposition, if the intersection of L with a regular language is
not context-free, then L was not context-free to begin with.  a*b*c*
is a regular expression, and the intersection of its language with L 
is a^n b^n c^n, which is not context-free.  So L is not context-free.