Dividend = Divisor $\times$ Quotient + Remainder
$397246 = x \times 865 + 211$
$397035 = x \times 865 \implies x = 397035 / 865 = 459$.

$(\sqrt{2})^2 = 2$, which is a rational number. Others contain irrational parts.

$\sqrt{-4}$ is an imaginary number.

There might be a typo in the question. If it is $36.34\bar{3} \div 7$, the answer would be $5.1919... = 5.1\bar{9}$.

Explanation - $4x3 + 984 = 13y7$. Divisibility rule of 11: difference of alternating digits is 0 or multiple of 11. $(7+3) - (y+1) = 0 \Rightarrow 10 - y - 1 = 0 \Rightarrow y=9$. Then $4x3 = 1397 - 984 = 413 \Rightarrow x=1$. $x+y = 10$.

By definition, non-terminating non-recurring decimals are irrational numbers.

$1921 + x(0.15) = 2449 \Rightarrow 0.15x = 528 \Rightarrow x = 3520$.

In 17065809, the place value of 7 is 7000000, the place value of 6 is 60000, and the place value of 9 is 9. Sum = 7000000 + 60000 + 9 = 7060009.

LCM of 7, 11, 13 = $7 \times 11 \times 13 = 1001$.
$3000 \div 1001 = 2$ times (2002), remainder = $3000 - 2002 = 998$.

Largest 2-digit prime = 97. Smallest 2-digit prime = 11. Difference = $97 - 11 = 86$.

Let the numbers be x and y. $x+y=15$, $\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{3}{10} \Rightarrow \frac{15}{xy} = \frac{3}{10} \Rightarrow xy = 50$. The numbers are 10 and 5, whose sum is 15 and product is 50.

Let the number be $10x+y$. $xy + (x+y) = 10x+y \Rightarrow xy = 9x \Rightarrow y=9$. So the unit digit is 9.

$\sqrt{a^2 + b^2} = 613 \implies a^2+b^2 = 613^2$. $(a+b)^2 = a^2+b^2+2ab = 613^2+2ab$. Assuming $a=612$, $b^2 = 613^2-612^2 = 1225 \implies b=35$. Thus $a+b = 612+35 = 647$.

$2.3\bar{4} = \frac{234-23}{90} = \frac{211}{90}$.

Let the numbers be $x$ and $x+2$. $\frac{1}{x} - \frac{1}{x+2} = \frac{2}{255} \Rightarrow \frac{2}{x(x+2)} = \frac{2}{255} \Rightarrow x^2+2x-255=0 \Rightarrow (x+17)(x-15)=0 \Rightarrow x=15$. The numbers are $15$ and $17$. Sum $15+17=32$.

$16.02 \times 0.001 = 0.01602$.

Largest 3-digit number = 999.
$999 \div 35 \implies$ Quotient = 28, Remainder = 19.
Therefore, required number = $999 - 19 = 980$.

Let the number be $x$.
$\frac{x}{6} - \frac{x}{9} = 100$
$\frac{3x-2x}{18} = 100 \implies \frac{x}{18} = 100 \implies x = 1800$.

$x + \frac{105+90+56}{126} = \frac{377}{126} \implies x + \frac{251}{126} = \frac{377}{126} \implies x = \frac{377-251}{126} = \frac{126}{126} = 1$.

If $x=1$, $1+1/1=2$. $1^{100} + 1/1^{100} = 1+1=2$.