If a number ends with n zeroes, its square will end with 2n zeroes. 
Here, 60 ends with one zero, so its square will end with 2 zeroes.
It is important to note that this stands true only for natural numbers (not decimals).

$\text{The square of a number will end in 1,}$$\text{if the digit in the units place is either}$$\text{1 or 9}.$
$\text{We know that}{1}^2=1\text{and}{9}^2=81.$
$\text{Among the given options, 161 has 1}\text{in its units place.}$
$\text{Hence,}(161{)}^2\text{will have 1 in its units}\text{place.}$

Perfect squares cannot have 2, 3, 7 or 8 in their unit's place.
In the given options only 1681 does not end with a number other than these.
The perfect squares of numbers ending in 1 and 9 have 1 at their unit's place. 
Thus, 1681 could be a perfect square of an integer ending with either 1 or 9.

The numbers 7, 2 and 3 do not appear in the unit place of any perfect square. So, ABC2 and PQR7 can not be perfect squares and XX1 is the only number which could be a perfect square.

$\sqrt{\sqrt{81}}$ 
$=\sqrt{\left(\sqrt{9\times 9}\right)}[\because 9\times 9=9^2=81]$
$=\sqrt{\left(9\right)}[\because \sqrt{9^2}=9]=\sqrt{(3\times 3)}[\because 3\times 3=3^2=9]=3$

Square of a perfect cube is another perfect cube.
Let us take a number, n.
Its cube is $n^3$. Square of this cube would be a perfect square given by ${\left(n^3\right)}^2$, which can also be written as ${\left(n^2\right)}^3$, a perfect cube. 

By division method,
$\begin{array}{cc}& 32\\ 3& \overline{10}\overline{24}\\ & 9\downarrow \\ 62& 124\\ & 124\\ & 0\end{array}$
$\therefore$ The square root of 1024 is 32.

The square root is the inverse operation of a square. 

A division is the inverse operation of multiplication, not of a square.

The square root of a perfect square is a natural number.
$\sqrt{196}$  =  $\sqrt{14\times 14}$  =  14

$\sqrt{256}$  =  $\sqrt{16\times 16}$  =  16

Whereas 156 and 176 are not the perfect squares.
Hence, 196 and 256 are perfect squares as they are squares of 14 and 16 respectively.

64 is the square of 8 which lies between 50 and 70. The squares of 7 and 9 are 49 and 81 respectively which lie outside this range. Therefore there is one perfect square between 50 and 70.