You are rolling a die three times, whereby $X$ denotes the random variable which counts the sum of the values of the dice. Let $A$ denote the event that $X$ is even.
Clearly, 1. would imply 2., but I don't really see how I can quickly check for independence here.
On
Clearly, $X$ and $A$ cannot be independent. To see that, let me spell out the technicalities.
We assume that $X$ is defined on the probability space $(\Omega, \mathcal{A}, \mathbb{P})$.
Then $X$ and $A$ are independent iff every set in the $\sigma$-algebra generated by $X$ is independent from $A$. Since $X$ maps to $\{1, \dots, 18\}$ the $\sigma$-algebra generated by $X$ is simply $$ \sigma(X) = \{X^{-1}(B): B \subseteq \{1, \dots, 18\}\}.$$ Hence, independence of $X$ and $A$ would require for every $B \subseteq \{1, \dots, 18\}$ that $$ \mathbb{P}(A \cap X^{-1}(B)) = \mathbb{P}(A) \mathbb{P}(X^{-1}(B)).$$ However, $A$ itself is in $\sigma(X)$ which would lead to $$ \mathbb{P}(A \cap A) = \mathbb{P}(A) \mathbb{P}(A). $$ But that is not possible.
Obviously $X$ and $A$ are not independent. If we know the value of $X$, we also know the value of $A$.
For the second part, there are $216$ possibilities for the three rolls. In $27$ of them all rolls are even, in $27$, all rolls are odd. In $81$ cases, two rolls are odd and one is even, and in $81$, two rolls are even and one is odd. In the first and third cases, $X$ is even, and in the second and fourth, $X$ is odd.
Compare the first and second cases. There is a one-to-one correspondence between the outcomes, by subtracting $1$ from each of the rolls, if they are all even. So the total of the case $1$ rolls is $81$ more than the total of the case $2$ rolls.
We can map case $3$ to case $4$ by adding $1$ to each of the odd rolls and subtracting $1$ from the even roll, so the total of the case $3$ rolls is $81$ less than the total of the case $4$ rolls.
That is, the sum of the rolls when $X$ is even is the same as the sum of the rolls when $X$ is odd, and equation $2$ holds.