Answer
a) $\color{blue}{0.5^{\large \frac{1}{\theta+1}}}$
b) $ \color{blue}{\frac{-1 + \sqrt{5}}{2}}$
Work Step by Step
From Def. 3.5.2 (p. 145), for a continuous random variable $Y$, the median value $m$ is such that $\displaystyle\int_{-\infty}^m f_Y(y)\ dy = 0.5$.
a)
$\begin{align*}
\int_{-\infty}^m f_Y(y)\ dy &= \int_0^m (\theta+1)y^\theta\ dy,\quad m\in (0,1), \theta \gt 0 \\
&= \left. (\theta+1)\cdot\dfrac{y^{\theta+1}}{\theta+1}\ \right\vert_0^m \\
&= m^{\theta+1} - 0^{\theta+1} \\
\int_{-\infty}^m f_Y(y)\ dy &= m^{\theta+1}, \quad m\in (0,1), \theta \gt 0 \\
\end{align*}$
Thus,
$\begin{align*}
& \int_{-\infty}^m f_Y(y)\ dy = 0.5, \quad m\in (0,1), \theta \gt 0 \\ \\
\implies & m^{\theta+1} = 0.5 \\
\implies & \color{blue}{m = 0.5^{\large \frac{1}{\theta+1}}}
\end{align*}$
b)
$\begin{align*}
\int_{-\infty}^m f_Y(y)\ dy &= \int_0^m \left(y+\frac{1}{2}\right)\ dy,\quad m\in (0,1)\\
&= \left[\ \frac{y^2}{2} + \frac{y}{2}\ \right]_0^m \\
&= \left(\ \frac{m^2}{2} + \frac{m}{2}\ \right) - \left(\ \frac{0^2}{2} + \frac{0}{2}\ \right) \\
\int_{-\infty}^m f_Y(y)\ dy &= \frac{m^2+m}{2}, \quad m\in (0,1) \\
\end{align*}$
Thus,
$\begin{align*}
& \int_{-\infty}^m f_Y(y)\ dy = 0.5, \quad m\in (0,1) \\
\implies & \frac{m^2+m}{2} = 0.5 \\
\implies & m^2+m = 1 \\
\implies & m^2+m-1=0 \\
\implies & m = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2(1)}, \qquad [\ \text{use the quadratic formula}\ ] \\
\implies & \color{blue}{m = \frac{-1 + \sqrt{5}}{2}},
\end{align*}$
where we have disregarded the negative root since $m\in (0,1).$
