Q2.
Consider f1(x)=f(x)-a and g1(x)=g(x)-a
Then if u draw a simple graph (by shifting the two curves downward by 'a') then you will see that f1(x)= -g1(x)
as both graphs will be symmetrical about x-axis.
So, f(x)-a = -[g(x)-a]
==> f(x)+g(x) = h(x) = 2a
So h(x) is neither one-one nor onto