Genetics: pedigree following a rare trait autosomal recessive
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.  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: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$