Answer: Option B. -> $$\frac{4}{3}$$
$$\eqalign{
& a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2} + {{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr
& = \frac{{2\left[ {{{\left( {\sqrt 5 } \right)}^2} + 1} \right]}}{{5 - 1}} \cr
& = \frac{{2\left( {5 + 1} \right)}}{4} \cr
& = 3 \cr
& a.b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = 1 \cr
& {\text{Put value in expression}} \cr
& \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr
& = \frac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a + b} \right)}^2} - 3ab}} \cr
& = \frac{{{3^2} - 1}}{{{3^2} - 3}} \cr
& = \frac{{9 - 1}}{{9 - 3}} \cr
& = \frac{4}{3} \cr} $$

Answer: Option B. -> $$\frac{{11}}{5}$$
$$\eqalign{
& {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = 2}}{{\text{5}}^{(3x - 4)}} \cr
& \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{\left( {{5^2}} \right)^{(3x - 4)}} \cr
& \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^2}^{(3x - 4)} \cr
& \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^{(6x - 8)}} \cr
& \Rightarrow x + 3 = 6x - 8 \cr
& \Rightarrow 5x = 11 \cr
& \Rightarrow x = \frac{{11}}{5} \cr} $$

Answer: Option D. -> $$\frac{7}{8}$$
$$\eqalign{
& \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr
& = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr
& = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr
& = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr
& {\text{ = 1}} - {2^{ - 3}} \cr
& {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr
& {\text{ = }}\frac{7}{8} \cr} $$

Answer: Option A. -> 4
(256)0.16 × (256)0.09 = (256)(0.16 + 0.09)
$$\eqalign{
& = {\left( {256} \right)^{0.25}} \cr
& = {\left( {256} \right)^{\frac{{25}}{{100}}}} \cr
& = {\left( {256} \right)^{\frac{1}{4}}} \cr
& = {\left( {{4^4}} \right)^{\frac{1}{4}}} \cr
& = {4^1} \cr
& = 4 \cr} $$

Answer: Option B. -> 10000
$$\eqalign{
& {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr
& = {10^{150 - 146}} \cr
& = {10^4} \cr
& = 10000 \cr} $$

Answer: Option B. -> 13
$$\eqalign{
& {\text{Let}}\,{\left( {25} \right)^{7.5}} \times {\left( 5 \right)^{2.5}} \div {\left( {125} \right)^{1.5}} = {5^x} \cr
& {\text{Then}},\,\frac{{{{\left( {{5^2}} \right)}^{7.5}} \times {{\left( 5 \right)}^{2.5}}}}{{{{\left( {{5^3}} \right)}^{1.5}}}} = {5^x} \cr
& \Rightarrow \frac{{{5^{\left( {2 \times 7.5} \right)}} \times {5^{2.5}}}}{{{5^{\left( {3 \times 1.5} \right)}}}} = {5^x} \cr
& \Rightarrow \frac{{{5^{15}} \times {5^{2.5}}}}{{{5^{4.5}}}} = {5^x} \cr
& \Rightarrow {5^x} = {5^{\left( {15 + 2.5 - 4.5} \right)}} \cr
& \Rightarrow {5^x} = {5^{13}} \cr
& \therefore x = 13 \cr} $$

Answer: Option B. -> 125
$$\eqalign{
& {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr
& = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr
& = {\left( {25} \right)^{\left( {3/2} \right)}} \cr
& = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr
& = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr
& = {5^3} \cr
& = 125 \cr} $$

Answer: Option B. -> 1
Given exp. =
$$ = \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}}$$
$$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}}$$
$$\eqalign{
& = \frac{{ {{x^a} + {x^b} + {x^c}} }}{{ {{x^a} + {x^b} + {x^c}} }} \cr
& = 1 \cr} $$

Answer: Option B. -> $$\sqrt 2 $$ - 2
$$\eqalign{
& \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr
& = 2\sqrt 2 - 2 - \sqrt 2 \cr
& = 2\sqrt 2 - \sqrt 2 - 2 \cr
& = \sqrt 2 - 2 \cr} $$

Answer: Option A. -> 1
$$\eqalign{
& {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr
& = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr
& = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr
& = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr
& = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr
& = {\left( {\frac{1}{4}} \right)^0} \cr
& = 1 \cr} $$