Question 1
Okay, so I'll go through my own process for you here step by step, moving down the tree. Here's an annotated version of your diagram with my own thoughts (it's been a while since I've done this but should hopefully be accurate):
Generation 2: As you realised, the trait is autosomal recessive so the female (II:2) has the genotype aa. The male is considered wild type unless informed otherwise, giving him a genotype of AA. I think you had worked all this out, but they are shown in red on the diagram.
Generation 3: Using the two genotypes of the parents from II (red), we know that all the progeny in generation III are carriers - i.e. Aa genotypes. This is indicated for III:6 in orangey-brown. I think you also worked this out successfully. However in your original chart, III:7 is noted as Aa. III:7 is from outside of the affected family and would therefore again be considered to have be wild type and therefore instead have the genotype AA - shown in purple.
Generation 4: In order to work out the potential genotypes of IV:1 we have to do a cross between the two parents III:6 and III:7 - which works out as follows:
| Male |
| A a |
|------------------------|
F | |
e A | AA Aa |
m | |
a | |
l A | AA Aa |
e | |
As you can see, there is a 50% chance of IV:1 being healthy AA and a 50% chance of being a carrier Aa - shown in green. 882
As IV:2 is affected, we know his genotype is aa (pink).
Generation V: Now for answer to your first question. If IV:1 had the AA genotype then the child could never be affected, reducing the odds to 50% immediately. Of this remaining 50%:
| Male |
| a a |
|------------------------|
F | |
e A | Aa Aa |
m | |
a | |
l a | aa aa |
e | |
50% of the offspring would be affected.
Therefore the probability of V:I being affected is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Question 2
The second question seems very odd to me. If IV:2 was unaffected in the literal sense of the word they would be wild-type AA. This would make the probability of them having an affected child as 0.
As this isn't one of your options, I assume they mean that IV:2 is a carrier (Aa). We are this time therefore crossing either AA or Aa (IV:1) with Aa (IV:2). Again, if IV:1 is heterozygous dominant (AA) none of her children will be affected. This means the odds are again reduced by a half. In the case that she is heterozygous Aa then the cross becomes:
| Male |
| A a |
|------------------------|
F | |
e A | AA Aa |
m | |
a | |
l a | Aa aa |
e | |
Giving 25% healthy, 50% carrier and 25% affected.
As you're looking for affected children: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ 882