EDIT : Thanks to Andre, he gave me a slap in the face right when I needed it, i.e. before my applied analysis exam tomorrow. XD I'll make this answer right before I bring anymore shame on me.
My ex-approach for the case $Y=5$ is indeed making everything more complicated. The more easy and non-complicated way to do this is just use the definition of the expectation :
$$
\mathbb E(X \, | \, Y = 5) = \sum_{n=1}^{\infty} \, n \, \mathbb P(X = n \, | \, Y = 5).
$$
Now we just need to compute those probabilities. We know that $P(X = 5 \, | \, Y = 5) = 0$, so that removes this term. For the first four terms, note that
$$
\mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^{n-1} \left( \frac 15 \right) = \frac{4^{n-1}}{5^n}, n=1, 2, 3, 4.
$$
because the condition $Y=5$ only means for $X$ that the first four rolls cannot take the value $5$, hence the value of those rolls become independent and uniformly distributed over $\{1,2,3,4,6\}$.
For the next rolls, $Y=5$ gives information on the first five rolls but no information on the ones after. Thus
$$
\mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^4 \left( \frac 56 \right)^{n-6} \left( \frac 16 \right).
$$
The $n-6$ stands for the number of rolls after the first five rolls which are not a 6 before you actually get your first $6$ (which gives the $1/6$ term). Therefore, the series we look at in the expectation can be computed, after the first 5 terms, as the derivative of a geometric series in $\left( \frac 56 \right)$. I don't wanna compute it right now because I am going to mess it up, I am definitively too tired for this.
This is a hard problem, but doable, even without partitioning the set or doing otherwise strange things you won't think of while taking Exam P. By definition we have $$E[X|Y=2]=\sum xPr(X|Y=2).$$ From here, you can use your reasoning skills to figure out what these probabilities. $$Pr(X=1|Y=2)=\frac 15,$$ because we have 5 equally likely possibilities. Since we're given $Y=2$, we DO NOT consider getting a 6 on the 1st try a possibility. $$Pr(X=2|Y=2)=0,$$ since you cannot have both a 5 and a 6 when you roll the die. $$Pr(X=3|Y=2)=\frac 45\cdot 1 \cdot \frac 16,$$ because you must not roll a 5 or 6 on the first roll, the second roll is given and happens with probability 1, and on the 3rd roll you roll a 5. This should be obvious, but these events are independent, so you multiply them. Finally consider
$$Pr(X=4|Y=2)=\frac 45 \cdot 1 \cdot \frac 56 \cdot \frac 16 ,$$
because you must not roll a 5 or 6 on the first roll, the second roll has a value 6 with probability 1, you must not roll a 5 on the 3rd roll, and on the 4th roll you roll a 5. You can keep reasoning in this way but you should see the pattern. Now putting everything together, we have $$E(X|Y=2)= \sum xPr(X|Y=2)= 1(\frac 15) + 3(\frac 45)(\frac 16) + 4(\frac 45)(\frac 56)(\frac 16) + ...,$$
which is the same thing as the following series:
$$\frac 15 + (\frac 45) \sum_{x=3}^{\infty}x(\frac 56)^{x-3}(\frac 16).$$
Now let $k=x-3$ (alternatively you can let $k=x-2$, but I prefer the former), we have
$$\frac 15 + (\frac 45) \sum_{k=0}^{\infty}(k+3)(\frac 56)^k (\frac 16)= \frac 15 + (\frac 45) \sum_{k=0}^{\infty}k (\frac 56)^k (\frac 16) + (\frac 45)3 \sum_{k=0}^{\infty} (\frac 56) (\frac 16).$$
You should see a geometric random variable with $p=\frac 16$. To the right of the final equality, the first sum is the expected value and the second sum is summing the probability mass function over the sample space, which evaluates to 1. Lastly, in this form, $EK = \frac {1-p}p$. If you instead let k=$x-2$, then in that form $EK = \frac 1p$. Either way you'll arrive at the same answer. Continuing in the way we proceeded, our final answer is:
$$ \frac 15 + \frac 45 \cdot \frac {1- \frac 16}{\frac 16} + \frac 45 \cdot 3 = 6.6 $$
Best Answer
$\require{cancel}$The best way to get this is to notice that $X\sim \text{Geom}(p = 1/6)$ on $\{1,2,3,\dotsc\}$, and in particular $$X|Y = 1 \overset d= 1+X$$ as A.S. commented. This means
$$E[X|Y=1] = E[1+X] = 1+6 = 7.$$
There are a number of other ways to get this. To do it your way, then $$P(X = k|Y = 1)$$ means that I tell you that you have taken one turn and got a 5, what is the probability of getting the first six on roll $k$.
So,
and in general $$P(X = k|Y = 1) = (5/6)^{k-2}(1/6),$$ for $k\geq 2, 0$ otherwise. Thus $$E[X|Y=1] = \sum_{k = 2}^\infty kP(X = k|Y=1) = \sum_{k=2}^\infty k(5/6)^{k-2}(1/6).$$ But we can simplify our work if we play around a little. \begin{align*} E[X|Y = 1]&= \sum_{k=2}^\infty kP(X=k|Y=1)\\ &=\sum_{k=2}^\infty k(1-p)^{k-2}p\\ &=\frac{1}{1-p}\left[\sum_{k=2}^\infty k(1-p)^{k-1}p\right]\\ &=\frac{1}{1-p}\left[-p+\sum_{k=1}^\infty k(1-p)^{k-1}p\right]\tag 1\\ &=\frac{1}{1-p}\left[-p+E[X]\right]\\ &=\frac{1}{5/6}\left[-\frac{1}{6}+\frac{1}{1/6}\right]\\ &=7, \end{align*} where in $(1)$, I recognize that the series is simply the expectation of a geometric distribution with $p = 1/6$ on $\{1,2,3,\dotsc\}$. Also, at $(1)$ you can use the series they gave you if you pull out $p$ from the series. I didn't bother. I like my way better.