This is the exact (non-race?) condition for students to collectively provide answers to the old exam. 
We are a couple of (benevolent) students who have come up with the following answers. 
In case you don't agree with us, please let us know!
Exam OS1 (1DT044) / OSPP (1DT096), 2018-08-22
Mixed concepts
 1 - K
 2 - T
 3 - R
 4 - I
 5 - N
 6 - M
 7 - J
 8 - C
 9 - G
10 - S

Module 1
1.1 B
1.2 B
1.3 B
1.4 D
1.5 D
1.6 C

Module 2
2.1 C
2.2 A
2.3 C
2.4 D
2.5 B
2.6 D (The dup2 syscall is used for redirection.)
2.7 B (The PCB is managed by the kernel.)
2.8 B
2.9 A

Module 3
3.1 D
3.2 D
3.3 A
3.4 
Avereage response time =
  0 + (2-1) + (11-4) + (8-7)     9
= --------------------------- = --- = 2,25
               4                 4

Avereage waiting time =
  (4-2 + 15-5) + (2-1 + 5-4) + (11-4 + 16-15) + (8-7 + 18-11)    30
= ----------------------------------------------------------- = ---- = 7,5
                                  4                               4
3.5 
Preemptive Shortest Job First:
---------------------------------------------------------------------------------
|   P1  |   P2  |     P3    |      P1       |       P4      |   P5  |     P4    |
|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20

Round-Robin, quantum = 3:
---------------------------------------------------------------------------------
|     P1    |   P2  |     P1    |     P3    |     P4    |     P4    |   P5  | P4|
|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|___|
0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20

Module 4
4.1 C
4.2 A
4.3 D
4.4 D
4.5 TRUE
    while
    key
    lock
4.6 
Fill in the need matrix (0.5 pt):
   | A | B | C | D |
T0 | 2 | 1 | 0 | 4 |
T1 | 4 | 2 | 2 | 0 |
T2 | 0 | 1 | 1 | 1 |
T3 | 2 | 1 | 3 | 1 |

Show each step of the Banker’s algorithm using the available matrix and the done and choice vectors (2 pt):
     |   Available   |
Step | A | B | C | D | Choice
  1  | 1 | 1 | 1 | 2 |   T2
  2  | 6 | 2 | 2 | 2 |   T1
  3  | 6 | 3 | 4 | 2 |   T3
  4  | 6 | 5 | 7 | 3 |   --
  5  | - | - | - | - |

According to the result of Banker’s algorithm from above, is the state safe (0.5 pt)?
 - No. T0 will not be able to run.
 The whole exercise is somewhat pointless, because you can see that the vector of totally available is 
  <A,B,C,D> = <8,5,8,5>
 but T0's request for resource D is 6, which exceeds what is totally available ...

Module 5
5.1 stack
    heap
    data
    text
5.2 C
5.3 8kB = 2^3 • 2^10 B = 2^13 B, so 13 bits are used for each page (offset).
    0x   A    6    1    9  =
    0b 1010 0110 0001 1001 =
    --  5 0 0110 0001 1001 which according to the page table translates to
    --  3 0 0110 0001 1001 =
    0x 0110 0110 0001 1001 =
    0x   6    6    1    9
    = = = = = = = = = = =

5.4 B
5.5 C