Pedigree Probability, Autosomal Recessive Trait
Here is my attempt!

Generation III


III3 is rR
II3 is rR and therefore III4 is rR with probability $\frac{1}{2}$ and RR with probability $\frac{1}{2}$.
Just like III4, III7 and III8 are rR with probability $\frac{1}{2}$.


Generation IV

Let's start with IV1.


IV1 is rR if ...


III3 transmits the r allele (probability $\frac{1}{2}$)
III4 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$). The total probability is $\frac{1}{4}$.

However, we have to remove the probability that both parent transmit the r allele as we know IV1 is not rr. This probability is $\frac{1}{2} \frac{1}{4} = \frac{1}{8}$. So, in total, the probability of IV1 to be rR is $\frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8}$.


Now, IV2


IV2 is rR if ...


III7 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$) or if III8 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$).

However, we have to remove the probability that IV2 is rr as we know this is not the case. The total probability is therefore $\frac{1}{4} \frac{1}{4} - \frac{1}{16} = \frac{7}{16}$.


Generation V

Finally, we get to your question


  [W]hat is the probability the offspring of IV1 x IV2 will show the recessive trait?


Both parents need to be rR which happen at probability $\frac{5}{8} \frac{7}{16} = \frac{35}{128}$ and both need to transmit the $r$ allele which happen with probability $\frac{1}{4}$. Hence the probability is $\frac{35}{512}$.