Rather than doing them for you, I'll guide you through problem number three and have you do the rest. Problem one may look tough, but it's rather simple if you know properties of exponents and logs.
Alrighty, problem number three:
\(x + y = -2\)
\(y+4x=x^2\)
So, using substitution, we want to get one of the equations to start as either y= or x=. I'll just pick the first equation and make it start with y=.
\(x +y = -2 \Rightarrow x + y -x = -2 - x \Rightarrow y = -2-x\)
So, now our system looks like this:
\(y=-2-x\)
\(y+4x=x^2\)
Now let's work on the second equation in our system. Since we know that y = -2 - x, we can just plug that in whenever we see a y in equation two:
\(y+4x = x^2 \Rightarrow (-2-x) + 4x = x^2 \)
Woohoo! All x's! Let's simplify this to get our x-values now.
\(-2-x+4x = x^2 \Rightarrow -2 + 3x = x^2 \Rightarrow 0 = x^2-3x+2\) -->Now we factor
\(0 = x^2 -3x +2 \Rightarrow 0 = (x-2)(x-1) \Rightarrow 0 = x-2, 0 = x-1 \Rightarrow x=2, x=1\)
So, we've got our two x-values! Neat! Now let's go back to equation 1 and find our y-values using those nifty x-values we just found.
\(y = -2 -x\Rightarrow y = -2 - 2, y = -2 - 1 \Rightarrow y= -4, y = -3\)
So, we've got two x-values and two y-values; now we just put them together and we've got two nice ordered pairs as answers!
Pair 1: (2, -4)
Pair 2: (1, -3)
The other 2 follow the same process as this, just some different algebra to find your values.
But, anyways, you could isolate pi from a number of formulas to isolate it and get an equation upon which you can solve for pi, such as pi = circumference / diameter. Hence, if you had an exact circumference and diameter, you'd likely be able to compute pi, but good luck with that. Pi is also equal to half of tau, which can be used in place of formulae such as C = 2 * pi * r, upon which this can be simplified to C = tau * r. 
