In $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, the denominator has to be rationalised for simplification. So , 
$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$\times$$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
$=\frac{(\sqrt{5}-\sqrt{3}{)}^2}{{\sqrt{5}}^2-{\sqrt{3}}^2}$
$=\frac{1}{2}$[${\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\sqrt{15}$]
$=4-\sqrt{15}$ 
$\text{So, a=4 and b=1.}$
$\therefore$a + b = 4 + 1 = 5

Irrational numbers are defined as those numbers which have a non-terminating and non-repeating decimal expansion and hence cannot be represented in the $\frac{p}{q}$ form. So, here we see that out of the given options the number 5.2731687143725186..... has a non-terminating and non-repeating decimal expansion as clearly one can interpret that there are no patterns formed in the decimal expansion and the number goes on expanding arbitrarily.

The integers include positive integers i.e. 1, 2, 3, ... , negative integers i.e. -1, -2, -3, ... and zero. Hence, zero is neither a positive nor a negative integer. 

A common tendency to solve this question is to apply the algebraic identity $(a^2-b^2)=(a+b)(a-b)$. But by observation, we can see that $(\sqrt{2}-\sqrt{2})$, which is one of the factors, will result in 0, thus making the final value of the whole expression as 0 i.e.
 $(\sqrt{2}+\sqrt{2})(\sqrt{2}-\sqrt{2})$
$=(\sqrt{2}+\sqrt{2})\times \left(0\right)$
$=0$

For any positive real number '$a$' and integers '$m$' and '$n$', we define
1. $a^m\times a^n=a^{m+n}$
2. ${\left(a^m\right)}^n=a^{mn}$
3. $(ab{)}^m=a^m{b}^m$  and if  $a=b$, the expression becomes  $(a\times a{)}^m=(a^2{)}^m=a^{2m}$.
​​​​​​So, using first rule, we have
${14}^{\frac{1}{7}+\frac{1}{7}}={14}^{\frac{2}{7}}$
Now, using second rule,
$({14}^2{)}^{\frac{1}{7}}$= ${14}^{\frac{2}{7}}$
Finally, using third rule,
($14\times 14{)}^{\frac{1}{7}}={14}^{\frac{2}{7}}$
Hence, we see that all the three results are same.

Let 'x' = 14.287628762876 ------- (i)
then 10000x = 142876.287628762876 --------(ii)
Subtracting (i) from (ii) we get
10000x - x = 142876.287628762876  - 14.287628762876
9999x = 142862
 then 'x' = $\frac{142862}{9999}$  
p = 142862, q = 9999
So, p-q = 132863

Natural numbers start from 1 and continue thereafter by adding 1 each time.

So 1, 2,3,4,5... are all natural numbers.

All the natural numbers and "0" together are referred to as whole numbers. Hence 1 is a whole number.

Natural numbers, their negatives and 0 constitute the set of Integers.

An irrational number is a real number that cannot be expressed as a ratio of integers, i.e. as a fraction, where the denominator is different from zero. Therefore, irrational numbers, when written as decimal numbers, do not terminate, nor do they repeat. $\sqrt{2},\sqrt{3}$ are examples of irrational numbers.

The decimal expansion of  $\frac{1}{4}$ is 0.25 and the decimal expansion of $\frac{1}{3}$ is 0.333...
Now comparing with the decimal expansion 0.25 and 0.333... with the options:
a) Clearly, 0 is out of the range.
b) The decimal expansion of $\frac{1}{8}$ is 0.125 which is also out of the range. 
c) The decimal expansion of $\frac{7}{25}$ is 0.28, so it lies between $\frac{1}{4}$ and $\frac{1}{3}$ .
d) The decimal expansion of $\frac{1}{5}$ is 0.2, so it is out of the range.

Let x = 1.6666......  ----- (i)
then, 10x = 16.666666 ----- (ii)
Subtracting (i) from (ii), we get
10x - x = (16.666666...) - (1.666666...)
$\Rightarrow$ 9x = 15
Hence x = $\frac{15}{9}=\frac{\frac{15}{3}}{\frac{9}{3}}=\frac{5}{3}$
i.e., p = 5 and q = 3
So, p + q = 8.