The size of the total space is $6^3$ distinct (and equally probable) outcomes.
Case 1) All faces the same: $X=1$. To obtain the size of the favoured space, count the ways to select one number to be repeated three times. As you did. $$\mathsf P(X{=}1)= \frac{6}{6^3} = \frac{1}{36}$$
Case 2) Two faces the same, one different: $X=2$. To obtain the size of the favoured space, count the ways to select a face for the pair, the ways to select a different face for the singleton, and the ways to choose which dice is a singleton. You weren't quite right.
$$\mathsf P(X{=}2)=\frac{6\times 5\times {^{3}{\rm C}_{1}}}{6^3} = \frac {5}{12}$$
Case 3) All faces distinct: $X=3$. Here we need to count the ways to select and arrange three distinct faces.
$$\mathsf P(X{=}2)=\frac{{^{6}{\rm C}_{3}}\times 3!}{6^3} = \frac {5}{9}$$
( Notice that the checksum is $1$. $\frac{1}{36}+\frac{5}{12}+\frac{5}{9}=1$ )
These are the probability mass function. $f_X(x)=\mathsf P(X{=}x)$
Now use these probabilities in the method you had.
$$\begin{align}\mathsf E(X) & = \sum_{x=1}^ 3 x\;\mathsf P(X{=}x)\\\mathsf {Var}(X) & = \sum_{x=1}^3 (x-\mathsf E(X))^2\mathsf P(X{=}x)\end{align}$$
alt: $\mathsf E(X^2) = \sum_{x=1}^3 x^2\,\mathsf P(X{=}x)$ and $\mathsf {Var}(X)=\mathsf E(X)^2 - \mathsf E^2(X)$
Let $p$ be the desired probability, and consider the first roll. It is either a $6$, in which case we're done and the sum is even, a $2$ or $4$, in which case we want the sum of the rest of the terms to be even, or a $1,3$, or $5$, in which case we want the sum of the rest to be odd.
Thus
$$p = \frac{1}{6}+ \frac{1}{3}p+\frac{1}{2}(1-p)$$
which simplifies to $p=\frac{4}{7}$.
Best Answer
It looks as if we are expected to assume the dice have numbers $1$, $2$, $3$ written on their faces. For the probability mass function of $X$, we need to calculate $\Pr(X=k)$ for all possible values of $k$. There values are $2$ to $6$.
Suppose one die is red and the other green. Record the result of a toss as an ordered pair $(u,v)$ where $u$ is the number on the green, and $v$ the number on the red. There are $9$ equally likely possibilities,
We get $X=2$ if we get $1$ on red and $1$ on gree. So $\Pr(X=2)=\frac{1}{9}$. We get $X=3$ if we have $1$ on green and $2$ on red, or the other way around. So $\Pr(X=3)=\frac{2}{9}$. Continue.
For $E(X)$, the mean of $X$, we need to calculate $$2\Pr(X=2)+3\Pr(X=3)+\cdots+6\Pr(X=6).$$
But there is a simpler way. Let random variable $U$ be the number on the green, and $V$ the number on the red. Then $X=U+V$ and therefore $E(X)=E(U)+E(V)$.
It is easy to see that $E(U)=E(V)=2$, so $E(X)=4$. We could also show that $E(X)=4$ by a symmetry argument.
For the variance of $X$, we can proceed from the definition of variance, or, somewhat more simply, from the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. Note that $$E(X^2)=2^2\Pr(X=2)+3^2\Pr(X=3)+\cdots+6^2\Pr(X=6).$$
A much nicer way is to note that since $U$b and $V$ are independent, the variance of $U+V$ is the sum of the variances of $U$ and $V$. These can be computed quickly.