\begin{aligned}
\frac{1}{1+a^{(n-m)}} + \frac{1}{1+a^{(m-n)}} = ?
\end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
= \frac{1}{\left( 1 + \frac{a^n}{a^m} \right)} +
\frac{1}{\left( 1 + \frac{a^m}{a^n} \right)} \\
= \frac{a^m}{(a^m+a^n)} + \frac{a^n}{(a^m+a^n)} \\
= \frac{(a^m+a^n)}{(a^m+a^n)} = 1
\end{aligned}

Answer: Option B

Explanation:

\begin{align}&\left(\dfrac{a}{b}\right)^{x-2} = \left(\dfrac{b}{a}\right)^{x-7}\\\\
&\Rightarrow \left(\dfrac{a}{b}\right)^{x-2} = \left(\dfrac{a}{b}\right)^{-(x-7)}\\\\
&\Rightarrow x - 2 = -(x - 7)\\\\
&\Rightarrow x - 2 = -x + 7\\\\
&\Rightarrow x-2 = -x + 7\\\\
&\Rightarrow 2x = 9\\\\
&\Rightarrow x = \dfrac{9}{2} = 4.5
\end{align}

Answer: Option C

Explanation:

\begin{aligned}
= \frac{(10^3)^7}{(10)^{18}}
\end{aligned}

\begin{aligned}
= \frac{(10)^{21}}{(10)^{18}} = 10^3 = 1000
\end{aligned}

Answer: Option D

Explanation:

Lets assume,
\begin{aligned}
\left(25 \right)^{7.5} \times \left(5 \right)^{2.5} \div \left(125 \right)^{1.5} = 5^x \\
\text{then, } \frac{ \left( 5^2 \right)^{7.5} \times \left(5 \right)^{2.5} }{\left(5^3 \right)^{1.5} } = 5^x \\
=> \text{then, } \frac{ \left( 5^{15} \right) \times \left(5^{2.5} \right) }{\left(5^{4.5} \right) } = 5^x \\

=> 5^x = 5^{15 + 2.5 - 4.5} \\
=> 5^x = 5^{13} \\
\text{Hence, } x = 13
\end{aligned}

Answer: Option B

Explanation:

\begin{aligned}
= 256^{0.16+0.09} = 256^{0.25} = 256^{\frac{25}{100}}
\end{aligned}

\begin{aligned}
= 256^{\frac{1}{4}}= (4^4)^{\frac{1}{4}}
\end{aligned}

\begin{aligned}
=(4)^{4 \times \frac{1}{4}} = 4
\end{aligned}