Given expression is
$4^3\times 4^{-2}\times {16}^2=4^3\times 4^{-2}\times {\left(\right(4{)}^2)}^2=4^3\times 4^{-2}\times 4^4(\because (a^m{)}^n=a^{mn})=4^{3-2+4}(\because a^m\times a^n=a^{m+n})=4^5$ 

$\because a^{-n}=\frac{1}{a^n}fora\ne 0$
$⟹(\frac{1}{5}{)}^{-3}=\frac{1}{(\frac{1}{5}{)}^3}=(\frac{5}{1}{)}^3=5^3=125$
Similarly, $(\frac{1}{3}{)}^{-4}=\frac{1}{(\frac{1}{3}{)}^4}=(\frac{3}{1}{)}^4=3^4=81$ and 
$(\frac{1}{4}{)}^{-3}=\frac{1}{(\frac{1}{4}{)}^3}=(\frac{4}{1}{)}^3=4^3=64$
$⟹\frac{(\frac{1}{5}{)}^{-3}-(\frac{1}{3}{)}^{-4}}{(\frac{1}{4}{)}^{-3}}=\frac{125-81}{64}=\frac{44}{64}=\frac{11}{16}$

$1,540,000,000=1.54\times 1000000000=1.54\times {10}^9$

The number written at the bottom is the base whereas the number at the superscript is known as the exponent. In the given example, 5 is the base and 3 is the exponent. 

3676.48 can be written in expanded form as:
$3\times {10}^3+6\times {10}^2+7\times {10}^1+6\times {10}^0+4\times {10}^{-1}+8\times {10}^{-2}$. 
 $a=3,b=6,c=1,d=-1,e=8a+b-c-d-e=3+6-1+1-8=1$

Given, $(2{)}^{m+1}\times 2^5=4^{-2}$
$\Rightarrow$ $(2{)}^{m+1}\times 2^5={\left(\right(2{)}^2)}^{-2}$
$\Rightarrow$ $(2{)}^{m+1+5}=(2{)}^{-4}$ 
Since the bases on  both the sides  are equal, therefore the exponents must also be equal.  This means, $m+1+5=-4$, or $m=-10$.

The total distance travelled by the spaceship between Earth and Saturn = Distance between (Earth & Mars + Mars & Saturn)
               2.2253x${10}^8$  kms + 1.19666 x ${10}^{9}kms$   
              = 2.2253x${10}^8$  kms + 11.9666 x ${10}^{8}kms$
                =14.1919 x${10}^8$ kms
                = 1.41919x${10}^9$$\times {10}^{3}m$ 
                = 1.41919x ${10}^{12}m=1419190000000m$       

$5.06\times {10}^{-4}=\frac{5.06}{10000}=0.000506$

${(2^5÷2^8)}^5\times 2^{-5}={\left(\frac{2^5}{2^8}\right)}^5\times 2^{-5}$
$={\left(2^{5-8}\right)}^5\times 2^{-5}(\because \frac{a^m}{a^n}=a^{m-n})$
$={\left(2^{-3}\right)}^5\times 2^{-5}$
$=2^{-15}\times 2^{-5}(\because (a^m{)}^n=a^{mn})$
$=2^{-15-5}(\because a^m\times a^n=a^{m+n})$
$=2^{-20}$
$=\frac{1}{2^{20}}(\because a^{-m}=\frac{1}{a^m})$

Note that $a^0=1$ for any non-zero integer $a$.
Also, for integers $a(\ne 0)$ and $m,$
$a^{-m}=\frac{1}{a^m}.$
$⟹4^{-2}=\frac{1}{4^2}=\frac{1}{16}$ and $4^{-1}=\frac{1}{4^1}$
 $\therefore 4^{-2}+4^{-1}+4^0+4^2=\frac{1}{16}+\frac{1}{4}+1+16$
$=\frac{1+4+16+256}{16}$
$=\frac{277}{16}$