(i)$(f \circ g)(x) = f(g(x)) =f( \log_b ( \frac{x+3} 2) + 1) = 2*b^{(\log_b ( \frac
{x+3} 2) + 1) -1} - 3 = (x+3) -3= x$.
(ii)$(g \circ f)(x) = g(f(x)) =g(2b^{x-1} -3) = \log_b ( \frac
{(2b^{x-1} -3)+3} 2) + 1 = (x-1)+1 = x $.
We check both with Mapping Diagrams.
Which do you find more convincing visually?