1-3. Solve matrix games:
1.

1'	3	5	7	1'	3	2
7*	8*	6*'	8*	7*	6*'	6*'
1	-5'	4	3	4	0	1
3	5	6*	-3'	0	1	3
2	3	-2'	0	1	2	0

* maximal in its column
' minimal in its row
There are 3 saddle ponts: the second row, columns 3, 6, and 7.
The value of game is 6.
The optimal strategy for the row player: choose the second row.
The optimal strategy for the column player: use any mixture of columns 3,6,7.

2.

0	0	0	-2	-2	-1	-2
0	4	3	-1	-2	-1	-1
0	4	3	0	-2	0	-1
3	0	-3	0	2	2	0
0	3	3	-2	-3	0	0

We name rows r1,r2,r3 and columns c1,c2,c3,c4,c5,c6,c7. By domination, we are reduced to the following 3 by 4 matrix game:

	c3	c4	c5	c7
r3	3	0	-2	-1
r4	-3	0	2	0
r5	3	-2	-3	0

It is easy to find good strategies for both players. To find optimal strategies, I used Mathematica 4.1 for Sun Solaris. Here is how I enter data:
 a = {{3, 0, -2, -1}, {-3, 0, 2, 0},  {3, -2, -3, 0}};
 x = {x3, x4, x5}; y = {y3, y4, y5, y7};
where
a is the payoff matrix,
x =p/(u+10) is   strategy for the row player p divided by (the value of game+10);
y =q/(v*10) is   strategy q for the row player divided by (the value of game+10).
To find optimal x:
 ConstrainedMin[Apply[Plus, x], Table[(x.(a + 10))[[i]] > 1, {i, 4}], x]
Answer:
{10/97, {x3 -> 3/97, x4 -> 11/194, x5 -> 3/194}
So the value of game is  97/10-10= -0.3
and an optimal strategy for the row player is  10x/97= [p3,p4,p5]^T=[0.3,0.55,0.15]^T.
Similarly, input
 ConstrainedMax[Apply[Plus, y], Table[((a + 1).y)[[i]] < 1, {i, 3}], y]
gives output
{10/7, {y3 -> 1/7, y4 -> 3/7, y5 -> 0, y7 -> 6/7}}
so an optimal strategy for the column player is
[q3,q4,q5,q7]=[0.1, 0.3, 0,0.6].
In the terms of the origional problem, optimal strategies are
[0, 0, 0.3,0.55,0.15]^T  and   [0, 0, 0.1, 0.3, 0, 0, 0.6].
 

3.

0	1	-1	1
-1	0	1	1
1	-1	0	1

The last column goes by domination. What is left is a familiar  symmetric game.
An optimal strategy for the row player is   [1/3,1/3,1/3]^T.
An optimal strategy for the column player is [1/3,1/3,1/3,0].
The value of game is 0.

4-5. Solve the linear programs, where  all  x_i>=  0:
4.

x1	x2	x3	x4	x5	1
1	2	-3	5	6	-2	= -x6
0	-1	2	0	3	1	= x7
-1	0	2	4	-4	2	= x8
2	0	3	0	1	2	= f -> min

We obtain a standard row tableau by multiplying the last column and the last  3 rows by -1:

x1	x2	x3	x4	x5	-1
1	2	-3	5	6	2	= -x6
0	1	-2	0	-3	1	= - x7
1	0	-2	-4	4	2	=  -x8
-2	0	-3	0	-1	2	= - f -> max

This is an optimal tableau, so the basic solution
x₁ = x₂ = x₃ = x₄ = x₅ = 0, x₆ =  2,   x₇=  1,  x₈ = 2,   max(-f)  =  -2, min(f) =2.
All optimal solutions are described as follows:
x₁ =   x₃ =  x₅ = 0,  x₆ = 2-  2x₁   -5 x₄ >=0,  x₇ = 1 -x₂ >=0, x₈=  2+ 4x₄, x₂ , x₄ >= 0.

5.

x1	x2	x3	x3	x5	-1
1	2	-3	5	6	-2	= -x6
0	-1	2	0	3	1	= x7
-1	0	2	4	-4	2	= 0
2	0	3	0	1	2	-> max

Combainning two  x₃ -columns into one and pivoting on 6 in this column gives a standard row tableau
with the first row bad. Also the first row in the origional tableau gives an infeasible constraint.
The program is infeasible.