Both problems CAN be done by long division.  I'll do the first
one by long division first just to show you that it can be done
that way.  Then I'll show you the shortcut, known as "synthetic"
division.  Here's the long division.  There is no x³ term so we
must put in a placeholder +0x³: 

            x³ + 2x² +  6x + 14
x - 2)x4 + 0x³ + 2x² +  2x +  5
      x4 - 2x³
           2x³ + 2x²
           2x³ - 4x²
                 6x² +  2x
                 6x² - 12x
                       14x +  5
                       14x - 28
                             33

The quotient is x³+2x²+6x+14 and the remainder is 33

However your teacher doesn't want you to do it that way. He/she wants
you to learn the shortcut, synthetic division, where you don't
have to write the variables down.  But synthetic division only works 
when you're dividing by either "x+↏" or "x-↏", so you can't do the 
second problem by synthetic division, but you can do the first one, 
that I just did above by long division.

In long division you have to subtract by thinking of the sign
changed and then add.  But in synthetic division, we change the
sign of the number after the x, so we can add instead of 
subtract.  Here we are dividing by x-2, so we change the sign 
of -2 to +2 and write 2 for the divisor.  Then instead of

1x4+0x³+2x²+2x+5 we just write 1  0  2  2  5 and draw a vertical
line between the 2 and the 1, and a horizontal line below like
this: 

2| 1  0  2  2  5
 |______________     

Begin by bringing down the 1

2| 1  0  2  2  5
 |______________
   1

Multiply the 1 by the 2 in the upper left, getting 2 and put it 
above and to the right, above the line underneath the 0:

2| 1  0  2  2  5
 |    2           
   1  

Now we add the 0 and the 2, getting 2, and write it below the
line under the 2:

2| 1  0  2  2  5
 |    2           
   1  2  

Multiply the 2 on the bottom line by the 2 in the upper left, 
getting 4 and put the 4 above and to the right, above the line
underneath the 2 in the middle on the top.

2| 1  0  2  2  5
 |    2  4      
   1  2 

Now we add the 2 and the 4 getting 6 and we write the 6 underneath
the 4 below the line:

2| 1  0  2  2  5
 |    2  4      
   1  2  6

Multiply the 6 on the bottom line by the 2 in the upper left, 
getting 12 and put the 12 above and to the right, above the line
underneath the next 2 on the top.

2| 1  0  2  2  5
 |    2  4 12   
   1  2  6

Now we add the 2 and the 12 getting 14 and we write the 14 underneath
the 12 below the line:

2| 1  0  2  2  5
 |    2  4 12   
   1  2  6 14

Multiply the 14 on the bottom line by the 2 in the upper left, 
getting 28 and put the 28 above and to the right, above the line
underneath the 5 on the top.

2| 1  0  2  2  5
 |    2  4 12 28
   1  2  6 14

Now we add the 5 and the 28 getting 33 and we write the 33 underneath
the 28 below the line:

2| 1  0  2  2  5
 |    2  4 12 28
   1  2  6 14 33
  
Finally we interpret the row of numbers  1  2  6 14 33 on 
the bottom.  Since the largest power of x in x^4-4x^2+2x+5 
is 4, the largest power of x in the quotient will be 1 
less than 4, or 3. So all the numbers but the last one are
the coefficients of the quotient, so the quotient is

1x³+2x²+6x+14 and the last number 33 is the remainder.

Notice that that is the same answer as when we used long division 
above.

--------------------------------------

We must put in a zero-placeholder in both the divisor and the dividend 
and consider this as dividing 2x5+4x4-x³-x²+0x+7 by 2x²+0x-1.   

This problem involves some zeros as well as some fractions at the end:

                          x³ + 2x² + 0x + 1/2 = quotient
2x² + 0x - 1)2x5 + 4x4 -  x³ -  x² + 0x +   7
             2x5 + 0x4 -  x³
                   4x4 + 0x³ -  x²
                   4x4 + 0x³ - 2x²
                         0x³ +  x² + 0x
                         0x³ + 0x² + 0x
                                x² + 0x +    7
                                x² + 0x -  1/2
                                     0x + 15/2 = remainder   

Notice I got the 15/2 remainder by 7- = 7+ =  = 

Edwin