for (1):

(1 + 2x)(1 - 2x) = 1 - 4x^{2}= 1 (since 4 = 0 (mod 4)).

for (2):

consider the homomorphism:

φ:(Z/4Z)[x]-->(Z/2Z)[x] given by:

a_{0}+ a_{1}x +...+ a_{n}x^{n}--> [a_{0}]+[a_{1}]x+...+[a_{n}]x^{n}, where [a_{k}] = a_{k}(mod 2).

if g(x) is a unit in (Z/4Z)[x], then g(x)k(x) = 1, whence φ(g(x))φ(k(x)) = φ(1) = 1, hence φ(g(x)) = φ(k(x)) = 1 (since the only units of (Z/2Z)[x] are the units of Z/2Z, that is: 1).

this means that φ(g(x) - 1) = φ(g(x)) - φ(1) = 1 - 1 = 0, that is g(x) - 1 is in ker(φ).

since the elements of ker(φ) are precisely the polynomials in (Z/4Z)[x] with coefficients of 0 or 2, g(x) - 1 = 2(f(x)), for some polynomial f(x).

hence g(x) = 1 + 2(f(x)).