Radial probability for orbitals
For the 1st one, the maximum probablity of finding the electron occurs at the Bohr's radius,i.e. at $\ce{r=r_0}$. This can be guessed by thinking this result classically. But if you want to know the exact method, the wave function for $1s $ orbital is, $$\ce{$\psi$_{1s} = 1/$\sqrt{\pi a_0^3}$ e^{-r/a_0}}$$ So, the radial probablity distribution for the $1s$ orbital will be,$$\ce{P(r) = |$\psi$|^2 4$\pi$r^2dr}$$ which will be eqal to , $$\ce{P(r) =4/a_0^3 r^2e^{-2r/a_0}}$$ To find the value of $r$ at which $\ce{P(r)}$ is maximum , you just have to find the solution of $\ce{dP/dr =0}$. You can verify very easily that the solution is $\ce{r=r_0}$.For the next one, the graph of B) is definitely for $2s$. Now, according to judgement of no. of radial nodes of $2s$ and $3p$, they  both have 1 radial node each. So, that is inconclusive. But if you would have known the wave functions, it would be easier.The wave-function of $2s$ is $$\ce{$\psi$_{2s} = c. (2- r/a_0)e^{-r/2a_0}, where c is a constant.}$$But, wave function of $3p$ orbitals are, $$\ce{$\psi$_{3p_z} = c_1(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}cos$\theta$}$$ $$\ce{$\psi$_{3p_x} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$cos$\phi$}$$ $$\ce{ ($\psi$_{3p_x} is actually an additive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$ and, $$\ce{$\psi$_{3p_y} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$sin$\phi$}$$$$\ce{ ($\psi$_{3p_y} is actually an subtractive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$So, you can see at $r= 0$, all the $3p$ orbitals have value of wavefunction (i.e. |$\psi$ |) = $0$ (simply check by putting $r$ as $0$ in all given $3p$ wave functions). But at $r =0$,$\ce{ $\psi$_{2s}}$ has non-zero finite value so, $\ce{|$\psi$|^2}$ does not vanish at origin for$ 2s$ orbitals which the given graph is showing. So, it can't be $3p$ as the value of the wave function doesn't match at origin. If you consider higher orbitals like $4d $ also, their wavefunction will also vanish at origin. So, the figure B) corresponds to $2s$.