1. The pedigree could show an AD, AR or XR mode of inheritance. It is most likely AD.  It could be AR if the mother was a carrier, and the father was a homozygote.  It could be XR if the mother was a carrier, and the father was a hemizygote. It cannot be XD, since the daughter (#2) would have necessarily inherited the disease allele on the X chromosome she received from her father.
2. There are many possible answers. Here are some possibilities:  If neither of the parents of the father were affected (i.e., the paternal grandparents of children 1, 2, 3), then the disease could not be dominant.  If only the paternal grandfather was affected, then the disease could only be X-linked recessive if the paternal grandmother was a heterozygote (which would be unlikely given that this is a rare disease allele).
1. The mode of inheritance is most likely AD, since every affected individual has an affected parent, and the disease is inherited even in four different matings to unrelated, unaffected individuals. It is very unlikely that it is XD or XR, in part because affected father had an affected son.
2. The mode of inheritance cannot be AD or XD, because affected individuals must have an affected parent when a disease allele is dominant. Neither can it be XR, because there is an affected daughter of a normal father.  Therefore, it must be AR, and this is consistent with the pedigree.
3. The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It is not XR, because there are unaffected sons of an affected mother.  It is therefore, likely, AR — but note that the recessive alleles for this condition appear to be relatively common in the population (note that two of the marriages were to unrelated, affected individuals).
4. The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It could be either XR or AR, but because all the affected individuals are male, and no affected males pass the disease to their sons, it is likely XR.
3. If a represents the disease allele, individuals a, d, f (who all married into this unusual family) are AA, while b, c, e, g, h, i, j are all Aa, and k is aa.
4. There is a ½ chance that an offspring of any mating Aa x AA, will be a carrier (Aa). So, there is a ½ chance that #3 will be Aa, and likewise for #4.  If #3 is a carrier, there is again a ½ chance that #5 will be a carrier, and likewise for #6.  If #5 and #6 are both Aa, then there is a ¼ chance that this monohybrid cross will result in #7 having the genotype aa and, therefore, affected by the disease.  Thus, the joint probability is $\frac{1}{2}\times\frac{1}{2} \times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{4} =\frac{1}{64}$.