For part 1), your answer is fine. I would get the same answer in
essentially the same way.
For part c), you have found the probability of correctly guessing the
red and blue dice while incorrectly guessing the green die.
But that is not the same thing as what we usually would mean by
"guessing correctly the outcome and the colour for exactly two dice".
You can fix the answer by considering how many different pairs of colors
there are whose outcomes could be guessed correctly.
For part d), you have found the probability of correctly guessing the
red die and incorrectly guessing the other two dice.
Similarly to part c), you can fix the answer by considering the number of colors that could have been the correctly guessed color.
I do not think there is a simple answer for part 2),
"guessing the outcome for three dice, but not all colours".
According to the problem statement, it seems to me the player is allowed
to guess "red $1$, blue $1$, green $1$".
In that case, either the player will get at least one number wrong
or the player will get all colors correct.
For that guess, the probability of
guessing the outcome for three dice but not all colors is zero.
On the other hand, if the player guesses "red $1$, blue $2$, green $3$"
then the outcome "red $1$, blue $3$, green $2$" has all three numbers correct
but only one color correct, while "red $3$, blue $1$, green $2$" has all three numbers correct but no color correct;
so for that guess, the probability to guess the outcome for three dice but not all colors is greater than zero.
To answer part 2) we need to make some assumptions about how the player
chooses which guess to make.
If we assume the player never mentions the same a number more than once
in a guess, suppose the player guesses "red $r$, blue $b$, green $g$"
where $r$, $b$, and $g$ are three particular elements of the set
$\{1,2,3,4,5,6\}$, no two of the numbers $r$, $b$, or $g$ being equal.
This is a correct guess of the "outcomes for the three dice"
if and only if when we read the numbers on the red, blue, and green dice
(in that sequence) after they have been rolled,
we get some permutation of the three numbers $r$, $b$, and $g$.
Each of these permutations is a member of the probability space for
rolling the three dice.
Now consider only the subset of these permutations for which the
player's guess does not have all three colors correct; this is exactly
the event "guessing the outcome for three dice, but not all colours".
Compute the probability of each of these permutations,
showing that each permutation has the same probability,
and multiply by the number of permutations in the event to obtain
the total probability of the event.
By the way, your answer for part 2) was equal to the probability of
guessing all three numbers and correctly guessing exactly one color;
but the method does not appear to be correct.
The probability of correctly guessing the outcome and color of the red die
(for example) is $1/6$.
Be aware that this is not how we would usually interpret "the probability of
guessing the outcome and color of one die," because guessing the outcome
and color of the blue die (or the green die) should also count as
guessing the outcome and color of one die.
But it is useful for the next part of the calculation to consider only
the event where the player correctly guesses the red die.
Then there are $36$ possible ways for numbers to appear on the blue
and green dice, but only one way for those numbers to match the two numbers of the player's guesses for the blue and green dice without
also matching the colors guessed by the player,
assuming the player did not guess the same number on both dice.
(If the player guessed the same number on each of those two dice, there is
of course no way for the dice to match the guessed numbers without also
matching the guessed colors.)
So given that the player correctly guessed the number and color of the red
die, and guessed two different numbers for the other two dice,
the chance to guess the numbers on those dice but not their colors is
$1/36$, not $1/12$.
The probability to guess the red die correctly and to guess the correct
numbers but not colors on the other two dice is therefore
$\frac16 \times \frac{1}{36} = 1/216$.
But that is only the probability to guess all the correct numbers
while guessing the correct color only on the red die.
We must also consider the cases where the player guesses the blue die
correctly, or the green die correctly.
If we assume that the player guesses a different number for each die,
each of the three cases has probability $1/216$ and
the total probability is $3 \times 1/216 = 1/72$.
This is still less than the probability that the player correctly guesses
all three numbers but not the correct colors,
because it does not count the event in which the player correctly
guesses the numbers but does not guess any colors correctly.
Best Answer
"Let's say we have one die then we have just 1 way to get at least one 6": yes, $1$ out of $6$ outcomes.
"For two dice, it becomes 6 ways". Nope, you have $1$ way to get a double 6, and $10$ ways to get only one 6 (first die: 6, second die: 1-5; first die= 1-5, second die: 6). $1+10=11$ out of $36$ outcomes.
"for three dice the answer comes to be 36 ways". Nope, you have $1$ way to get three 6s, $15$ ways to get two 6s (three times five ways: 6, 6, 1-5; 6, 1-5, 6; 1-5, 6, 6), $75$ ways to get only one 6 (three times twentyfive ways: 6, 1-5, 1-5; 1-5, 6, 1-5; 1-5, 1-5, 6). $1+15+75=91$ out of $216$ outcomes.
Hence even not using the (easier) not rule the probability of getting at least one 6 is $\frac{91}{216}$ :)