1. Let A₁ = C([-1,[( -1)/ 2]]È[[ 1/ 2],1]) and A₂ = C([0,1]È{2}). Since [0,1]È{2} isn't homeomorphic to [-1,[( -1)/ 2]]È[[ 1/ 2],1], A₁ isn't isomorphic to A₂. Also the function which sends x Î Inv(A₁) to y Î G₂ defined by

y(t) = ì ï ï ï ï ï ï ï í ï ï ï ï ï ï ï î x(t-1) t Î [0, 1 2 ] é ê ë x( -1 2 )/x( 1 2 ) ù ú û x(t) t Î [ 1 2 ,1] x( 1 2 )/x( -1 2 ) t = 2

is the desired isomorphism.
Ref.
[Zel] W. Zelazko, Banach algebras, Elsevier Publishing Company, 1973.