$\large N = 612182430\cdots$

1.

Since $N$ is very large number that is formed by the concatenation of multiples of $6$ . Counting all the multiples of $6$ is tedious to find the $1000^{th}$ digit in the large number $N$ .

In the number above $6$ is the one digit number in $N$ . The two digit numbers that are the mutiples of $6$ in between $10$ and $100$ initiate from $6\times2=12$ and terminates at $6\times16 = 96$ . Altogether there are $16-1 = 15$ (excluding $6$ being one digit number) two digits numbers and $15\times 2 = 30+1 = 31$ digits in the concatenation from $1$ to $100$ that are multiple of $6$ . Three digits numbers between $100$ and $1000$ that are mutiples of $6$ are $166-16 = 150$ (excluding 1 one digit and 15 two digits number). Total digits are $150\times 3+31 = 481$ in large number $N$ . To meet the $1000^{th}$ digit in $N$ , remaining digits are $1000-481 = 519$ which are formed by the concatenation of four digits numbers. Here, $519\equiv 3\mod(4)$ where $129$ are pairs of four digits numbers. These $129$ pairs of four digit numbers lies in between $167$ and $300$ . Since $519$ (remaining pairs of 4 digits) is not divisible by $4$ however, $520=1001-481$ or $516=997-481$ is divisible by $4$ which implies that $997^{th}$ and $1001^{th}$ digit will be $295\times 6 = 1770$ and $296\times 6 = 1776$ respectively. So $1000^{th}$ digit left hand side of $\large N= 612182430\cdots17701776$ will be $7$ .

2.

Total one digit number that is multiple of $6$ be $w$ . Then, $w = \left\lfloor \frac{10}{6}\right\rfloor = 1\times1 = 1$ Total two digits numbers $x =\left\lfloor\frac{100}{6}\right\rfloor-1 = 16-1 = 15 \Rightarrow 2x= 2\times15= 30$

Total three digits numbers $y = \left\lfloor\frac{1000}{6}\right\rfloor -16 = 150 \Rightarrow 3y = 450$

Concatenation of one, two and three digits that are multiple of $6$ are $1+30+450=481$ .

Since there are huge numbers of four digits between $1000$ and $10000$ digit numbers so such a mutiple of $6$ should be obtaines very close to $1000^{th}$ and to meet $1000^{th}$ digit in the above concatenation required numbers of are $1000-481= 519$ that total four digits numbers ie $4z =519 \Rightarrow z = \left\lfloor\frac{519}{4}\right\rfloor = 129$ pairs of four digits numbers that start from $167 = 167\times 6 = 1002$ to $296\times 6 = 1776$ that represents a pair of $4$ from $997^{th}$ to $1001^{th}$ digits in concatenation of multiples of $6$ .

Therefore, $1000^{th}$ digits is $7$ .