How do I do question 1c and 3? I'm having a hard time understanding punnet squares and all this genotype stuff, is there any really clear explanations anyone would recommend? Thankyou!!
and question 10 if anyone is keen, not to worried about it though, more after the theory behind it thanks 
Firstly, you didn't do question 1a quite right. Where there are two alleles at one gene locus, one recessive and one dominant, the dominant is a capital letter (e.g. B) and the recessive is lower-case of the same letter (e.g. b). It doesn't matter what letter you choose, but they MUST be the same.
So, let's call this the b gene. Which is dominant? Since we have a white and a brown, which produce ALL brown offspring, the brown must be dominant, because the white allele never shows through in the offspring.
--> brown = B, white = b.
What are the genotypes of the mice that were crossed? This means, what are the genotypes of the parents, the white parent and the brown parent. You know that each mouse has two copies of the gene (one on each of their chromosomes in the pair), one from their mum and one from their dad.
Now the white mouse MUST have both 'white' (b) alleles, because we already said that the 'brown' (B) allele is dominant; if the white mouse had the B allele, that would 'overcome' the b allele so the mouse would actually be brown. So the white mouse has the genotype
bb.
The brown mouse must have at least one B allele (because it's brown). BUT. If the other allele was 'white' (b), then 1/2 of its gametes would get the b allele, so 1/2 of its offspring would end up bb (one b from the brown parent, and one from the white parent), so 1/2 the offspring would be white. But since we have ALL offspring brown, the other allele must also be B, because every single offspring must have a B allele. ==> brown mouse genotype:
BB.
1c: So, each of their kids (F1 generation) have the genotype
Bb, because they get one B from the brown and one b from the white. So now you're crossing a Bb x Bb. Each parent has two possible gametes they pass down, so the punnett square shows the possible options of the next generation depending on which gametes they're 'made of'.
B bB BB Bb
b Bb bb
From this we see that 3/4 of the 2nd generation have the B allele, so 75% of that generation will be brown.
Question 3Let's go with: far-sighted = F, normal = f.
A normal man must have both f, because if he had the dominant F, then that would 'overcome' the f and he'd be far-sighted. ==>
ff.
The far-sighted women must have one F (because she's far-sighted), and her father was normal so he must have been ff, so she must get an f from him. ==>
Ff f fF Ff Ff
f ff ff
1/2 of those (50%) are ff, and thus normal.
Q10The mother's genotype must be
bb, because she's white and thus can't have a B allele. The father must have at least one B, because otherwise he wouldn't be black.
7 black, 6 white suggests about 50/50 in the offspring. Now if the father had BOTH B alleles, then every one of the offspring would get a B allele, so every one of them would be black. This isn't the case, so the father must be [/b]Bb[/b].
Well that was a long explanation. Let me know if anything doesn't make sense.