E and Z configuration
You are correct; the answer in your book is wrong. I get (E) for both alkenes. 

For the alkene on the upper right:


$\ce{COOH}$ has precedence over $\ce{CH2OH}$ because the first carbon atom is bonded to $\mathrm{ \{O,O,O\}}$ and the second is bonded to $\mathrm{\{O,H,H\}}$.
Clockwise around the ring has precedence over counterclockwise. While both initially are $\mathrm{\{C,H,H\}}$, the next carbon atom clockwise is $\mathrm{\{C,C,H\}}$ while the next carbon atom counterclockwise is $\mathrm{\{C,H,H\}}$.


For the alkene on the lower left, the nitrile groups win in both cases:


On the left: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{N,H,H\}}$
On the right: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{C,C,H\}}$