1)

x=0，g(x)=a

X≠0, g(x)=f(x)/x, f(0)=0, g(x) can be governed by L'H?pital's law:

lim(x→0) g(x)

=lim(x→0) f(x)/x

=lim(x→0) f'(x)/ 1

=f'(0)

=0

G(x) is continuous, then g(0)=a=0.

So: a=0

2)

x≠0，g'(x)=f'(x)/x-f(x)/x？ =[xf'(x)-f(x)]/x？

lim(x→0) g'(x)

=lim(x→0) [xf'(x)-f(x)]/x？ Apply L'H?pital's law

= lim(x→0)[f '(x)+xf ' '(x)-f '(x)]/(2x)

=lim(x→0) f''(x)/2

=f''(0)/2

So: g(x) has a first-order continuous derivative.