We have 68 < F < 77   ....(1)
Putting \(F=\frac { 9 }{ 5 } \) C +32 (1) becomes
 \(68<\frac { 9 }{ 5 } C+32,<77\)
⇒ \(68-32<\frac { 9 }{ 5 } C<77-32\)
⇒ \(36<\frac { 9 }{ 5 } C<45\)
⇒\(\frac { 5 }{ 9 } \times 36\)
⇒ 20 < C < 25
Range of temperature is between 20^oC and 25^oC

We have |x| ≤ k    ..(1)
Case I : When x ≥ 0
(1) ⇒ |x| ≤ k [∵ |x| = x]
 0 ≤ x ≤ k ⇒ x ∈ [0, k]
Case II : When x < 0
(1)  ⇒ -x ≤ k [∵ |x| = -x]
⇒ x ≥ k [∵ x ≤ y ⇒-x ≥ -y]
⇒  -k ≤  x
⇒ -k ≤  x ⇒ x ∈ [-k, 0]
∴ The solution is [-k, 0] ∪ [0, k] which is same as [-k, k]

Given inequation is
⇒ 5x-1 < 15
⇒ 5x < 15 + 1
⇒ 5x < 16
⇒ \(x < \frac { 16 }{ 5 } \)
⇒ x < 3.2
Since x ∈ N, we have x = 1, 2, 3 ...
∴ Solution set is {1, 2, 3}

= \({ (27) }^{ \frac { -2 }{ 3 } +\frac { 1 }{ 3 } }=({ 27 })^{ \frac { -1 }{ 3 } }={ 3 }^{ 3\times \frac { -1 }{ 3 } }={ 3 }^{ -1 }=\frac { 1 }{ 3 } \) \(\left[ \because \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n } \right] \)

= \({ 3 }^{ -6\times \frac { 1 }{ 3 } }={ 3 }^{ 2 }=\frac { 1 }{ { 3 }^{ 2 } } =\frac { 1 }{ 9 } \)

= \(\left( { -10 }^{ 3\times \frac { -2 }{ 3 } } \right) =-{ 10 }^{ -2 }=\frac { -1 }{ { 10 }^{ 2 } } =\frac { -1 }{ 100 } \)

= \(({ 2 }^{ 4 })^{ -\frac { 3 }{ 4 } ={ 2 }^{ 4\times \frac { 2 }{ 4 } }={ 2 }^{ -3 } }=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 } \)

\(\left( 5^{ 3 } \right) ^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }\) 
= 5² = 25  \(\left[ \because ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)

Given quadratic equation is \(\sqrt [ 8 ]{{{x}\over{x+3}} } -\sqrt{{{x+3}\over{x}}}=2.\)
Let y \(=\sqrt{x\over x+3}\Rightarrow{1\over y}=\sqrt{x+3\over x}\)
∴ (1) becomes 8y - \(\frac{1}{y}\) = 2

\(⇒\ {8y^2-1\over y}=2⇒8y^2-1=2\)
8y²-2y-1 = 0
(2y - 1)(4y + 1) = 0
2y = 1 or 4y = -1
\(y={1\over 2}\)or \(y={-1\over 4}\)
\(\sqrt{x\over x+3}={1\over 2}\sqrt{x\over x+3}={1\over 2}\ or\ \sqrt{x\over x+3}={-1\over 4}\)
Case(i) \(\sqrt{x\over x+3}={1\over 2}\Rightarrow{x\over x+3}={-1\over 4}\)
4x = x + 3 ⇒ 3x = 3 ⇒ x = 1
Case(ii) \(\sqrt{x\over x+3}={-1\over 4}\)
This is impossible since LHS is non-negative.
∴ The root is 1.

Given \(\frac { 1 }{ \left| 2x-1 \right| } <6\)

Multiplying the numerator and denominator by |2x-1| we get, \({|2-1|\over |2x-1|^2}<6\)
⇒ |2x-1| < 6|2x-1|²
⇒ 1< 6 |2x-1|
\(⇒\ {1\over 6}<|2x-1|\)
\(⇒\ |2x-1|> {1\over 6}\)
\(⇒\ {-1\over 6}\ge2x-1\ge{1\over 6}\)
\(⇒\ {-1\over 6}+1\ge2x\ge{1\over 6}+1\)
\(⇒\ {5\over 6}\ge2x\ge{7\over6}\)
\(⇒\ {5\over12}\ge x\ge {7\over 12}\)
∴ The solution set is \(\left( -\infty,{5\over 12}\cup [ {7\over 12},\infty\right)\)

Suppose √3 is a rational number
Then √3 can be written as \(√3={m\over n}\)
Where m and n are rational numbers with no common factors other than 1.
Squaring both sides we get,
\(3={m^2\over n^2}⇒3n^2=m^2\)
multiplying by 2 we get
6n² = 2m² ⇒ 3(2n²) = 2 m²
Since 2n² is divisible by 2, m² is also an even number
⇒ m must be even
⇒ m = 2k for some natural number k
⇒ 3n² = (2k)²
⇒ 3n² = 4k²
⇒ n is also an even number
Thus both m and n are even numbers having a common factor.
This contradicts our initial assumption that m and n do not have a common factor
Hence √3 cannot be a rational number.
⇒ √3 is an irrational number.
Hence proved.

Given quadratic equation is
5^2x-5^x+3 + 125 = 5x
\(\Rightarrow\) (5^x)²-5^x 5³- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-125 5^x- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-126.5^x+125=0
Let 5^x=y
\(\Rightarrow\)y²-126y + 125 = 0
\(\Rightarrow\)(y-1)(y-125) = 0
\(\Rightarrow\)    y = 1 or 125
\(\Rightarrow\)   5^x = 1 or 125
Case (i) When 5^x = 1 \(\Rightarrow\) 5⁰ \(\Rightarrow\) x = 0
Case (ii) When 5^x = 125 \(\Rightarrow\) 5^x = 5³\(\Rightarrow\) x = 3
\(\therefore\) The roots are 0, 3.

Given 4 - x \(\le\) 3x + 12
\(\Rightarrow\) -x \(\le\) 3x + 12 - 4
\(\Rightarrow\) -x \(\le\) 3x + 8
\(\Rightarrow\) 3x + 8 \(\ge\) -x     \([x\le y\Rightarrow y \ge x]\)
\(\Rightarrow\) 4x + 8 \(\ge\) 0
\(\Rightarrow\) 4x \(\ge\) -8
\(\Rightarrow\) x \(\ge\) \({{-8}\over{4}}\)
\(\Rightarrow\)x \(\ge\) -2
\(\therefore\) x \(\in\) \([-2,\infty).\)
\(\therefore\) The solution set is  \([-2,\infty).\)

Given inequality is \({{2x+5}\over{x-1}}>5\)
\(\Rightarrow\) \({{2x+5}\over{x-1}}-5>0\)
\(\Rightarrow\) \({{2x+5-5x+5}\over{x-1}}>0\)
\(\Rightarrow\) \({{-3x+10}\over{x-1}}>0\)
Case(i) -3x + 10 > 0, x - 1 > 0
-3x < -10, x >1
3x < 10, x >1
\(x<{{10}\over{3}},x>1\)
\(1,x<{{10}\over{3}}\)
Case(ii) -3x + 10 < 0, x - 1< 0
-3x < -10, x < 1
3x<10,  x <1
\(x>{{10}\over{3}},x<1\)
This is impossible
\(\therefore\)  The solution is \(\left(1,{{10}\over{3}} \right)\)

Given \(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \)   ..(1)
Multiplying each term by the conjugate of the denominator we get
 \(\frac { 1 }{ 3-\sqrt { 8 } } \) = \(\frac { 1 }{ 3-\sqrt { 8 } } \times \frac { 3+\sqrt { 8 } }{ 3+\sqrt { 8 } } =\frac { 3+\sqrt { 8 } }{ { 3 }^{ 2 }-\sqrt { 8 } ^{ 2 } } =\frac { 3+\sqrt { 8 } }{ 9-8 } =3+\sqrt { 8 } \)
\(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \)= \(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \times \frac { \sqrt { 8 } +\sqrt { 7 } }{ \sqrt { 8 } +\sqrt { 7 } } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 8-7 } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 1 } =\sqrt { 8 } +\sqrt { 7 } \)
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \) =\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } +\sqrt { 6 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } ^{ 2 }+\sqrt { 6 } ^{ 2 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\sqrt { 7 } +\sqrt { 6 } \)
\(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \)  = \(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \times \frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } +\sqrt { 5 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } ^{ 2 }+\sqrt { 5 } ^{ 2 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ 6-5 } \)
 \(\frac { 1 }{ \sqrt { 5 } -2 } \) =  \(\frac { 1 }{ \sqrt { 5 } -2 } \times \frac { \sqrt { 5 } +2 }{ \sqrt { 5 } +2 } =\frac { \sqrt { 5 } +2 }{ \sqrt { 5 } ^{ 2 }+2^{ 2 } } =\frac { \sqrt { 5 } +2 }{ 5-4 } =\sqrt { 5 } +2\)
Substituting all these values in (1)we get
\(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \) = 5

\(5 x-1 <24 \ \ \ 5 x+1 >-24 \)
\(5 x <25 \ \ 5 x >-25\)
\(x <5 \ x >-5 \)
\(x \in(-5,5)\)

\(\text { In }(-\infty, 2), \frac{|x-2|}{x-2} \text { is negative }\)
\(\operatorname{In}[2, \infty), \frac{|x-2|}{x-2} \geq 0\)

\(|x+2| \leq 9 \)
\( \Rightarrow-9 \leq x+2 \leq 9 \)
\( \Rightarrow-11 \leq x \leq 7 \)
\(x \text { belongs to }[-11,7]\)