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Set the r's equal  and we have that

2cos θ = 1       divide both sides by 2

cos  θ  =  1/2

arccos (1/2) =  θ

Remember that arccos  has a range of  [-pi,2, pi/2 ]

Note that this happens   at  pi/3    and at  -pi/3 =  5pi/3       if we are on the interval [0, 2pi ]

So...the intersection points   are

  (r,  θ) =   (  1, pi/3)   and (1, 5pi/3 )      

First one

Notice that the top figure is a parallelogram with a base of 6 and a height of 4

So its area = 6 * 4 = 24 units^2

And the bottom figure is a triangle with a base of 6 and a height of 2

So its area is (1/2)(6)(2) = 6 units^2

So...the total area is 4 + 6 =  30  units^2

Last one

f(x) is the same  on both graphs

If we put 0 into 2^x  we get 2^(0)  = 1

So...the point (0, 1) is on the graph of  g(x)

So...again....the top graph is correct.....and the intersection points are at 0 and 2

Second one

If we put   0 into f(x), we get 2

Thus the point (0, 2)  is on the graph of f(x)

So....the first graph must be correct

The intersection point  is just  at   0

Since both of these equations describe height.....we are looking for the time when they are equal....so set them equal to each other and solve for x...this will be the time when they are at THE SAME HEIGHT...

-4.9x^2+12x+2.5  =  -4.9x^2 +14x

12x+2.5 = 14x

2.5 = 2x

x = 2.5/2 = 1.25 seconds

Not done yet......this is the TIME when they are at the same height....we need to sub this value of x into one of the equations to find the height at this time

-4.9(1.25)^2 + 14 (1.25) = ~~  9.8 m

First one

g(x)  is the same on both graphs

f(x) = 1/x - 2      is shown on the top graph   

So....the top graph is correct

The intersection points are  - 1  and 1

P.S. - If you allowed to......Desmos is a pretty good graphing utility that might help you

You can find it here :   https://www.desmos.com/calculator

Under the 'x' column....scroll down to 1   look across....the value under  f(x) = g(x)     this is ONE point......do you see others?

  try it....