Answer
a) $6.5\;\rm eV$
b) $276\;\rm nm,\;355\;nm$
c) Ultraviolet.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the ionization energy is the energy required to remove an electron completely from the atom, meaning to take it from its lowest energy state (ground state) to the point where it is no longer bound to the atom.
From the given sketch, it is obvious that the ionization energy is given by
$$|E_1|=0-(-6.5)=\color{red}{\bf 6.5}\;\rm eV$$
$$\color{blue}{\bf [b]}$$
For the absorption spectrum, we know that the energy absorbed by the atom is given by
$$E_{\rm photon}=\Delta E_{\rm atom}=\dfrac{hc}{\lambda}$$
Hence, the wavelengths are given by
$$\lambda=\dfrac{hc}{\Delta E_{\rm atom}}$$
$$\lambda=\dfrac{hc}{E_f-E_i}\tag 1$$
From the given sketch, it is obvious that the absorption spectrum consists of the transitions from the ground state, $1\rightarrow 2$ and $1\rightarrow 3$.
Using (1) to find both;
$$\lambda_{1\rightarrow 3}=\dfrac{hc}{E_3-E_1} =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(-2+6.5)(1.6\times 10^{-19})}=\color{red}{\bf 276} \;\rm nm$$
$$\lambda_{1\rightarrow 2}=\dfrac{hc}{E_2-E_1} =\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(-3+6.5)(1.6\times 10^{-19})}=\color{red}{\bf 355} \;\rm nm$$
$$\color{blue}{\bf [b]}$$
Since both values above are less than 400 nm, both wavelengths are ultraviolet.
