Answer
(a) ${\bf 0.342}\; a_B^{-3/2}$
(b) ${\bf 0.368 }\; a_B^{-1}$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The radial wave function for the hydrogen atom in the 1s state is given by
$$
R_{1s}(r) = \frac{1}{\sqrt{\pi a_B^3}} e^{-r/a_B}\tag 1
$$
So at $ r = \frac{1}{2} a_B $,
$$
R_{1s}\left(\frac{1}{2} a_B\right) = \frac{1}{\sqrt{\pi a_B^3}} e^{-\frac{1}{2}}
$$
$$
R_{1s}\left(\frac{1}{2} a_B\right) =\color{red}{\bf 0.342}\; a_B^{-3/2}
$$
$$\color{blue}{\bf [b]}$$
The probability density $ P_r(r) $ is given by
$$
P_r(r) = 4\pi r^2 |R_{nl}(r)|^2
$$
For $ r = \frac{1}{2} a_B $, we get:
$$
P_{1s}\left(\frac{1}{2} a_B\right) = 4\pi \left(\frac{a_B}{2}\right)^2 \left(\frac{1}{\sqrt{\pi a_B^3}} e^{-\frac{1}{2}}\right)^2
$$
$$
P_{1s}\left(\frac{1}{2} a_B\right) =\color{red}{\bf 0.368 }\; a_B^{-1}
$$
