I have 12 spots and three colors

%100 because there are more than 3 colours

like 6%

this lol

We have the event: "Some color will be missing", Call this Event X

Partition this event into the following events:
A) "Only red appears"
B) "Only green appears"
C) "Only yellow appears"
D) "At least 1 red and at least 1 green appears, but no yellow"
E) "At least 1 red and at least 1 yellow appears, but no green"
F) "At least 1 green and at least 1 yellow appears, but no red"

Pr(X) = Pr(A) + Pr(B) + Pr(C) + Pr(D) + Pr(E) + Pr(F)

Pr(A), Pr(B), Pr(C) are easy to compute
Pr(D), Pr(E), Pr(F) are not as obvious to compute

Let's also define the following events, whose probabilities are easy to compute:
G) "No yellow appears"
H) "No green appears"
I) "No red appears"

Pr(D) = Pr(G) - Pr(A) - Pr(B)
Pr(E) = Pr(H) - Pr(A) - Pr(C)
Pr(F) = Pr(I) - Pr(B) - Pr(C)

I'll leave the rest of the work for you to do

You're not including TWO colours missing as satisfying the condition "A colour will be missing"?
That's what the 'At least 1 x and at least one y, but no z' assures, right?

If you are referring to the events G, H, and I as "A colour will be missing", those events, as defined, do indeed include the cases where two colors are missing. These events were defined simply because their probabilities are easy to compute. They are not directly used to compute the probability of event X, they are used to compute the probability of events D, E, and F.

The events D, E, and F are worded in a way to specifically exclude the cases where two colors are missing, which is required to create a valid partition of X, in other words, to avoid double counting.

Pr(X) = Pr(A) + Pr(B) + Pr(C) + Pr(D) + Pr(E) + Pr(F)

Pr(D) = Pr(G) - Pr(A) - Pr(B)
Pr(E) = Pr(H) - Pr(A) - Pr(C)
Pr(F) = Pr(I) - Pr(B) - Pr(C)

So subbing the G -A -B etc. bits into the original sum, we get

Pr(A) + Pr(B) +Pr(C) + Pr(G) - Pr(A) -Pr(B) +Pr(H) - Pr(A) -Pr(C) +Pr(I) -Pr(B) -Pr(C)

= Pr(G) - Pr(C) + Pr(H) - Pr(B) + Pr(I) - Pr(A)

Now Pr(G) - Pr(C) is
Prob. of no yellows - prob. of only yellows

and Pr(H) - Pr(B) is prob.of no greens - prob.of only greens

and also prob.of no reds - prob.of all reds

Why are we subtracting the probabilities of all one colour, when that DOES mean there's colours missing?

The way you are grouping and interpreting the final expression is not intuitive and natural, you need to consider the expression as a whole. Consider the following:

The full event space can be partitioned into the following events:
Ev1: “0 red, 0 green, 1+ yellow” - same as only yellows showing up
Ev2: “0 red, 1+ green, 0 yellow” - same as only greens showing up
Ev3: “1+ red, 0 green, 0 yellow” - same as only reds showing up
Ev4: “0 red, 1+ green, 1+ yellow”
Ev5: “1+ red, 0 green, 1+ yellow”
Ev6: “1+ red, 1+ green, 0 yellow”
Ev7: “1+ red, 1+ green, 1+ yellow”
Note “0 red, 0 green, 0 yellow” is technically an event, but it has probability = 0, so I will ignore it.
You should be able to convince yourself that these events do indeed form a valid partition.

Event X can be partitioned into Ev1 through Ev6 inclusive

Event G can be partitioned into Ev2, Ev3, Ev6
Event H can be partitioned into Ev1, Ev3, Ev5
Event I can be partitioned into Ev1, Ev2, Ev4
Note that the above three events include Ev1 through Ev6 inclusive but Ev1 through Ev3 inclusive are double counted. This is compensated by considering the events:

Event A is the same as Ev3
Event B is the same as Ev2
Event C is the same as Ev1

When you consider the probabilities of such events, you can see that plugging in,

Pr(X) = Pr(G) - Pr(C) + Pr(H) - Pr(B) + Pr(I) - Pr(A)
can be represented as
Pr(X) = Pr(Ev1) + Pr(Ev2) + Pr(Ev3) + Pr(Ev4) + Pr(Ev5) + Pr(Ev6)

75% not red
75% not green
50% not yellow

would be 2* 0.75^12 + 0.5^12 ?

12 "spots"
25% red
25% green
50% yellow

red missing:
25% red
75% non-red
= (0.75)^12

green:
25% green
75% non-green
= (0.75)^12

yellow:
50% yellow
50% non-yellow
= (0.5)^12

add up possibilities of all non-red, all non-green and all non-yellow:
2 (0.75)^12 + (0.5)^12 ~= 0,0633 + 0,0002
~= 0,0635

Each of those calculations includes the cases of only one colour present, causing these cases to be included twice in the calculation. For example, P(All Yellow) is a case that is included in both P(No Red) and P(No Green).
You would need to subtract each of these cases once to accommodate for this.
As such, proper sum is:
P(No Green) + P(No Yellow) + P(No Red) - P(All Green) - P(All Yellow) - P(All Red)
2(0.75)^12 + 0.5^12 - 2(0.25)^12 - 0.5^12
Which is around 0.0634

This is just a special case of the problem solved here

Ah, of course. This best explains

>Why are we subtracting the probabilities of all one colour

The answer is around 6.3% as simulated on python. I don't know where these guys are getting 0.0634 from, they are orders of magnitude wrong. Math brainlets BTFO

That's because probabilities go from 0 to 1, the answer is the same. Although to be fair if OP wrote his probabilities in percentages then the answers should have been written in the same way.