Explanation

(a)
$f^{'}(5) = \lim_{h \rightarrow 0}{\ \frac{f(5 + h) - f(5)}{h}}\ \$
$= \ \lim_{h \rightarrow 0}\ \ \frac{2f(5)f(h) - f(5)}{h}$
$= \ \lim_{h \rightarrow 0}\ \ 2f(5)\lbrack\frac{f(h) - \frac{1}{2}}{h}\rbrack$
$\Rightarrow \ 1024\log{2 = 2f(5)f^{'}(0)}$
Again now, $f(2 + 3) = 2f(2)f(3)\ \ \ldots(i)\$
$\Rightarrow \ \ \ \ \ \frac{1024\log 2}{2f^{'}(0)} = 2 \times 8 \times f(3)$
$\Rightarrow \ \ \$ $f(3) = \frac{32\log 2}{f^{'}(0)}$ …(ii)
$\therefore\ \ f^{'}(3) = \lim_{h \rightarrow 0}\ log\ \frac{f(3 + h) - f(3)}{h}$
$= \lim_{h \rightarrow 0}\ \frac{2f(3)f(h) - f(3)}{h}$
$= 2f(3)f^{'}(0)$
$= 2 \times \frac{32\log{2f^{'}(0)}}{f^{'}(0)}$
=64 log 2 [from Eq. (ii)]