1)a)i) in book
1)a)ii) in book
1)a)iii) in book

1)b) in book

1)c)i) in book
1)c)ii) in book
1)c)iii) in book
1)c)iv) in book

2)a)i) forward
2)a)ii) reverse 
2)a)iii) reverse 
2)a)iv) forward
2)b) in book
2)c)i) in book
2)c)ii) in book
2)c)iii) in book
2)c)iv) in book

3)a) in book
3)b) in book

4)a)i) in book
4)a)ii) I believe it would be close to 0V as the input of 100Hz is very low, and this is a high pass filter meaning only high frequency signals should pass through
4)b) They can have a gain greater than 1, they can be stricter (have a sharper cut off, higher Q factor), than a passive filter
4)c)i) in book
4)c)ii) in book

5) 
- the diode must be between the mesurment middle point and point C, this is because there is 0 current flowing when C is connected to 9V
- you can also tell this because current flows from A-B and B-A just fine
- there is a resistor going from A to the mid point, and from B to the mid point
- the resistor from A to the mid point must be 100 ohms, this is because there R = V/I, we can use the result from A-C (resistor through diode) to tell that V = 9 - 0.7 and the table to tell us that I = 0.083 which gives us 100 ohms
- the resistor from B to the mid point must be 100 ohms, this is because there R = V/I, we can use the result from B-C (resistor through diode) to tell that V = 9 - 0.7 and the table to tell us that I = 0.0415 which gives us 200 ohms
- this can be confirmed by the fact that using the table values fro A-B or B-A, we can say 9/0.3 to get the resistance or Ra + Rb which is 300 ohms, thus confirming our result from before