Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L),...

Step 1
Each of three successive vehicles can go in three directions, denote
R - vehicle turns right on a particular freeway exit;
L - vehicle turns left on a particular freeway exit;
S - vehicle goes straight on a particular freeway exit.
For example, event [RLS] would mean that the first vehicle turned right, second vehicle turned left and third vehicle went straight.
a): All possible outcomes of all three vehicles go in the same direction are
${\displaystyle A=\{RRR,LLL,SSS\}}$.
Step 2
b): All possible outcomes of all three vehicles take different direction are
${\displaystyle B=\{RLS,RSL,LRS,LSR,SRL,SLR\}}$,
as we can see, the order of letters is important.
c): All possible outcomes of exactly two of the three vehicles turn right means that we need exactly two out of three letters to be "R"
${\displaystyle C=\{RRL,RRS,LRR,SRR,RSR,RLR\}}$.
Step 3
d): All possible outcomes of exactly two vehicles go in the same direction is
${\displaystyle D=\{RRL,RRS,LRR,SRR,RSR,RLR,LLR,LLS,RLL,SLL,LRL,SSL,SSR,SRS,SLS,LSS,RSS\}}$
Here, every went should contain two of the same latter (any of the three) and the third letter should be different (any of the remaining two) from the one we took in the first step, this way we obtain that exactly two vehicles go in the same direction.
e): D' is the complement of event D. The complement (opposite) of D means that all vehicles go in the same direction or they go in different direction (to avoid that exactly two vehicles go in the same direction).
Therefore ${\displaystyle D^{\prime }=\{RRR,LLL,SSS,RLS,RSL,LRS,LSR,SRL,SLR\}}$.
${\displaystyle C\cup D}$ is event C union event D. Event C is subset of event D (meaning that exactly two vehicles turn right is "subset" of exactly two vehicles go in the same direction). This indicates that ${\displaystyle C\cup D=D}$.
For the same reason we have that ${\displaystyle C\cap D}$(C intersection D) is
${\displaystyle C\cap D=C}$.

Answer:
a. ${\displaystyle A=\{RRR,LLL,SSS\}}$
b. ${\displaystyle B=\{RLS,RSL,LSR,LRS,SRL,SLR\}}$
c. ${\displaystyle C=\{RRS,RRL,RSR,RLR,LRR,SRR\}}$
d. ${\displaystyle D=\{RRS,RRL,RSR,RLR,LRR,SRR,SSR,SSL,SRS,SLS,LSS,RSS,LLS,LLR,LRL,LSL,SLL,RLL\}}$
e. ${\displaystyle C\cup D=\{RRS,RRL,RSR,RLR,LRR,SRR,SSR,SSL,SRS,SLS,LSS,RSS,LLS,LLR,LRL,LSL,SLL,RLL\}}$
${\displaystyle C\cap D=\{RRS,RRL,RSR,RLR,LRR,SRR\}}$
Step-by-step explanation:
a. All three vehicles go in the same direction
This means that all the three vehicles either go straight, or right or left at the same time.
So, ${\displaystyle A\{\{RRR,LLL,SSS\}}$
b. All three vehicles take different directions
Thus means that all the vehicle take different routes
If a vehicle takes the the straight direction, the other 2 vehicles take right and left directions respectively
So, ${\displaystyle B=\{RLS,RSL,LSR,LRS,SRL,SLR\}}$
c. Exactly two of the three vehicles turn right
This means that two vehicles take right direction while the last vehicle either tske the straight route or left route.
${\displaystyle C=\{RRS,RRL,RSR,RLR,LRR,SRR\}}$
d. Exactly two vehicles go in the same direction.
The means that if two vehicles pass straight direction then the third vehicle takes either the right or left direction.
Also, if two vehicles pass right direction, the third vehicle either takes the left direction or straight direction
Lastly, if two vehicles pass left direction then the third vehicle either pass the straight direction or right direction.
${\displaystyle D=\{RRS,RRL,RSR,RLR,LRR,SRR,SSR,SSL,SRS,SLS,LSS,RSS,LLS,LLR,LRL,LSL,SLL,RLL\}}$
e. ${\displaystyle C\cup D=}$ the union of items in C and D without repetition
So,