Need solving these 4 continuous functions/limits problems.
even with just one would be appreciated.

Please only post one problem at a time. I'll answer problem 1 to get you started. If you still need help with the rest, then post them as separate posts and I'll have a look at them. 

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Notice how x = -1 is the junction point, where one piece transitions to the other

Plug x = -1 into each piece of the piecewise function

So for the first piece we have
b - 2x
b - 2(-1)
b + 2

and similarly for the second piece
-6/(x-b)
-6/(-1-b)

Now set the two expressions equal to each other and solve for b
-6/(-1-b) = b+2
-6 = (b+2)(-1-b)
(b+2)(-1-b) = -6
-b-b^2-2-2b = -6
-b-b^2-2-2b+6 = 0
-b^2-3b+4 = 0
b^2+3b-4 = 0
(b-1)(b+4) = 0
b-1 = 0 or b+4 = 0
b = 1 or b = -4

The value of b with greater absolute value is b = -4 as this value is further from zero (compared to b = 1)
So -4 goes in the box

This is what the graph of f looks like (see attached) when b = -4

Answer

1 number

explanation

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

 

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.