Assessing probability from a pedigree
II:6 has a $2/3$ probability of being a carrier Aa.
II:3 has a $1/2$ probability of being a carrier Aa. 
Assuming, just like canadianer
suggested, that the recessive gene is rare, autosomal, and
non-x-linked, this means II:5 and II:4 are both homozygous (AA).


If parents of III:2 are therefore Aa and AA, this gives III:2 the probability of being a carrier of $1/2$. However, we must multiply this value by $1/2$, because III:2's father has a $1/2$ probability of being Aa. Therefore $\frac{1}{2}\frac{1}{2}= 1/4 =$ probability of III:2 being a carrier. 

Similarly for III:3, if her parents are Aa and AA, this gives III:3 the probability of being a carrier of $1/2$. However, we must multiple this value by $2/3$, because III:3's mother has a $2/3$ probability of being Aa. Therefore $\frac{1}{2}\frac{2}{3}= 1/3 =$ probability of III:3 being a carrier. 

The probability of both III:2 and III:3 being carriers is the product of both their probabilities: $\frac{1}{3}\frac{1}{4}= 1/12$.