Note both the answer obtained by following your solution and the answer in the book agree. According to your solution the answer should be $$\dfrac{1}{2}\cdot\dfrac{1}{25} + \dfrac{1}{2}\left(\dfrac{1}{25} + \dfrac{24}{25}\cdot\dfrac{1}{25}\right) = \dfrac{37}{625}$$ and the solution in the book is $$\dfrac{1}{2}\cdot\dfrac{1}{25} + \dfrac{1}{2} \left(1 - \left(\dfrac{24}{25}\right)^2\right) = \dfrac{37}{625}.$$ Note here, the probability of "not losing both dart throws" is $1 - \left(\dfrac{24}{25}\right)^2$.

For part (b) we need to divide the probability of getting heads and winning by the probability of winning. Thus, we want to divide $\dfrac{1}{2}\cdot\dfrac{1}{25}$ by $\dfrac{1}{2}\cdot\dfrac{1}{25} + \dfrac{1}{2} \left(1 - \left(\dfrac{24}{25}\right)^2\right)$.