The question is not clear. I propose to interpret it thusly.
You're given a subset $L$ of $\{{1,2,3,4,5,6,7\}}$. You're also given digits $a$ and $b$. You want to know whether there's a number $Y$ that starts with $a$, ends in $b$, has all its other digits from $L$, and is divisible by 3; then you want to know the same, but divisible by 7. If that's not the intended interpretation, maybe OP will return to let us know.
For divisibility by 3, if the number ab (by which I mean $10a+b$, not $a\times b$) is a multiple of 3, then no matter what $L$ is, you can let $Y={\rm ab}$. If the number ab is not a multiple of 3, then you can find $Y$ if and only if $L$ contains at least one of the digits 1, 2, 4, 5, 7. If $d$ is any such digit, then either adb or addb will be a multiple of 3.
Divisibility by 7 is a bit trickier. For example, if $a=3$, $b=0$, $L=\{{1\}}$, then you're out of luck, since all the numbers 30, 310, 3110, 31110, etc., leave remainder 2 on division by 7. Another example is $a=7$, $b=1$, $L=\{{7\}}$, when you're looking at the numbers 71, 771, 7771, 77771, etc. I think it shouldn't be hard to characterize the cases where there is no such $Y$, and I think that if $L$ has more than one element there are no such cases, but I leave the details as exercises.
EDIT: a couple more cases. $a=7$, $b=1$, $L=\{{3\}}$ can't be done; if $Y=7333\dots331$ then $3Y+7=22000\dots000$ which is not a multiple of 7, so $Y$ is not a multiple of 7. If $L$ contains any number other than 3 or 7, we win: 7111111, 721, 7441, 75551, and 766661 are all multiples of 7. We also win if $L=\{{3,7\}}$, since 7371 is a multiple of 7. This problem is more interesting than I thought it would be, and it might be fun for someone to do it right, that is, for every possible $a,b$ determine the sets $L$ that do, or don't, admit a value of $Y$.