Q:

Unsure of meaning of assignment function (variable assignment) in semantics of predicate logic?

I'm currently in a mathematical linguistics course, and I'm having trouble understanding the meaning of 
'g[d/v]: the variable assignment g′ that is exactly like g except (maybe) for
g(v), which equals the individual d'
in the semantics of predicate logic. If given a variable assignment in an example model, does this mean that (v) refers to all variables (d) that is in the universe, thus that all elements of the universe replaces the variable (v)?
g1 =
x1 → John
x2 → Mary
x3 → Pete
xn → Pete
(where n≥4)
g1[John/x3] =
x1 → John
x2 → Mary
x3 → John
xn → Pete (where n≥4)
g1[[John/x3]Pete/x1] =
x1 → Pete
x2 → Mary
x3 → John
xn → Pete (where n≥4)
Also, I have an exercise based on variable assignment equivalence,  but I do not know how to approach answering these questions since I do not entirely understand the meaning of variable assignment, and modified variable assignment. 
QUESTION: Complete the equivalences assuming: g(x) = Mary, and g(y) = Susan.
1.
g[Paul/x)(x) =
2.
g[Paul/x)(y) =
3.
g[[Paul/x]Susan/x)(x) =
4.
g[[Paul/x]Susan/x)(y) =
5.
g[(Paul/x)Susan/y)(x) =
6.
g[[Paul/x]Susan/y)(y) =
If anyone could explain this concept to me, I would be very grateful!
EDIT: sorry, i'm quite new to this site! the questions were cut off by the closed bracket. I've tried attempting the questions below.
1.g[Paul/x)(x) = x: Paul
2.g[Paul/x)(y) = y: Susan 
3.g[[Paul/x]Susan/x)(x) = x: Susan? 
4.g[[Paul/x]Susan/x)(y) = y: Susan? 
5.g[[Paul/x]Susan/y)(x) = x: Paul? 
6.g[[Paul/x]Susan/y)(y) = y: Susan? 
I'm a bit unsure about some of these, if x is originally mapped to Mary, then to Paul & Susan in (3&4)

A:

If given a variable assignment in an example model, does this mean that (v) refers to all variables (d) that is in the universe, thus that all elements of the universe replaces the variable (v)?

No.  A variable assignment maps every variable to a specific individual. You can see that with your first example:

g1 =
x1 → John
x2 → Mary
x3 → Pete
xn → Pete (where n≥4)

However, we can change those assignments when we do something like:

g1[John/x3]

This means that everything gets assigned the same individual as above, except that we now map $x3$ to John, so we get:

g1[John/x3] =
x1 → John
x2 → Mary
x3 → John
xn → Pete (where n≥4)

So, for last exercise at the end, your initial g is:
g  =
x → Mary
y → Susan
So that means that g[Paul/x] is:
g[Paul/x] =
x → Paul
y → Susan
Can you do the others?

