Well, since I'm new around here and you asked so niceley, here are some semi-solutions (I don't like giving the whole game away, it spoils all the fun )
1. A) Turn the equations into their matrix form:
(1 -6 2 5)
(2 -9 -1 14)
(4 12 -3 19)
and use elementary row operations (e.g. start with Row2 -> Row2 - 2*Row1, etc) to reduce to echelon form. You should then have a last row like (0 0 <something> <something else>), which means you have a simple equation for z. use this value in the middle row to get y, and this value in the first row to get x and you've got your solution.
C) When you do the row operations, if one of the rows is some integer multiple of another, you have infinitely many solutions (since they are essentially the same equation, so you have three variables and two equations, leaving one free variable that can take any value it likes).
Use one of the rows with two variables to obtain an equation for one in terms of the other (e.g. y = 4z +2) and solve the other equation using this value. What you should end up with is two equations that give two of the variables in terms of the other (e.g. x = 2z +4, y = 5z - 7).
D) If you do some row operations, you should find that one of the rows becomes something like 0x + 0y + 0z = <something>.
2. A) Let f(x) = x3 - 3x2 + 2x +1. Now, if you can show that f(x) must cross the x-axis between x=-1 and x=0, it must have a solution (a point where f(x) = 0) between -1 and 0. Newton's method is just number crunching once you know what numbers to crunch.
B) z-w and zw are just as they look, remember when you work out z/w, it's always a good idea to multiply the numerator and denominator by the complex conjugate of the denominator, in this case -1-2i.
D) If one solution is (-1-5i), then another will be (-1+5i), the complex conjugate of the first. You then have two factors of the function, namely (z + 1 - 5i) and (z + 1 + 5i). Multiply them together and find what other term you need to use to get to your original answer, and you're home and dry.
If you need any more help, feel free to ask.