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.
Each element in the sample space can be represented by the ordered triple $(b, g, r)$, where $b$ represents the outcome on the blue die, $g$ represents the outcome on the green die, and $r$ represents the outcome on the red die. Since $b, g, r \in \{1, 2, 3, 4, 5, 6\}$, there are six possible values for each of the three entries. Therefore, the size of the sample space is $6^3 = 216$.
In the comments, you asked why we use different colors for the dice. If we used three white dice, there would be
$$\binom{6}{1} + \binom{2}{1}\binom{6}{2} + \binom{6}{3}$$
distinguishable outcomes. The first term represents the number of outcomes in which all three dice show the same value. The second term represents the number of outcomes in which two dice show one value and the other die shows a different value. The factor of $\binom{6}{2}$ represents the two values shown by the three dice. The factor $\binom{2}{1}$ represents the two ways one of those two values can appear twice. The third term represents the number of outcomes in which the three dice show different values.
We do not use this sample space since the outcomes are not equally likely to occur. Each of the six outcomes in which all three dice show the same value can occur in only one way. However, each of the $30$ outcomes in which two of the dice show one value and the other die shows a different value can occur in three ways and each of the $20$ outcomes in which the dice show three different values can occur in $3! = 6$ ways. Notice that
$$\binom{6}{1} + 3 \cdot \binom{2}{1}\binom{6}{2} + 6 \cdot \binom{6}{3} = 216$$
As for the probability that exactly two sixes occur when the three different color dice are shown, observe that there are $\binom{3}{2}$ ways to choose which two of the three distinguishable dice show a $6$ and $5$ ways for the third die to show a value other than $6$. Hence, the number of favorable cases is
$$\binom{3}{2} \cdot 5$$
so the desired probability is
$$\frac{\binom{3}{2} \cdot 5}{6^3} = \frac{3 \cdot 5}{216} = \frac{5}{72}$$
If you instead meant at least two sixes, we gain the additional favorable outcome that three sixes occur, in which case the probability is
$$\frac{\binom{3}{2} \cdot 5 + 1}{6^3} = \frac{3 \cdot 5 + 1}{216} = \frac{16}{216} = \frac{2}{27}$$
Best Answer
Your version has two errors.
$(1)\;$You didn't choose which die is the solo one. To fix it, you need a factor of $3$.
$(2)\;$Of your two factors of $5$, the second one is incorrect. It should be $1$, not $5$, since there is no choice of color for the third die.
With those corrections, you get the product $$(3)(6)(5)(1) = 90$$ which matches the official answer.