What proportion of the population are carriers/heterozygous of the Albino allele?

This is a Hardy-Weinberg thing isn't it? To be honest, it's a long time since I did this, so you'd better check it:

It's (A + a) or equivalently (A + a)^2 = 1 (total population)
which is A^2 (homozygous dominant) + 2Aa (heterozygous) + a^2 (albino) = 1
you have the frequency of albinoism: 1/20,000 (ie 0.00005)
the frequency of the the 'a' allele is the square root of that: 0.00707 (3 significant figures).  The frequency of the A allele is 1 - 0.00707 = 0.993.

Frequency of heterozygous (Aa) = 2 x 0.993 x 0.00707 = 0.0140 (again to 3sf)