Answer
See the detailed answer.
Work Step by Step
a) In example 32-2, the cyclotron accelerates protons but here we need to accelerate alpha particles.
This means that the charge is doubled and the mass is now 4 times that of the proton.
Thus, the kinetic energy is given by
$$KE=\dfrac{q^2B^2R^2}{2m}=\dfrac{(2e)^2B^2R^2}{2m_{\alpha}}$$
Plugging the known;
$$KE=\dfrac{(2\times 1.6\times 10^{-19})^2(1.7)^2(0.25)^2}{2(4\times 1.66\times 10^{-27})}=\bf1.39277\times10^{-12}\;\rm J $$
$$ KE=1.39277×10^{-12}\;\rm J\times \dfrac{1\;eV} {1.6\times 10^{-19}\;J} =\color{red}{\bf 8.70482}\;\rm MeV $$
The velocity is given by
$$KE=\frac{1}{2}m_{\alpha}v^2$$
Thus,
$$v=\sqrt{\dfrac{2KE}{m_\alpha}}=\sqrt{\dfrac{2\times 1.39277×10^{-12}}{4\times 1.66\times 10^{-27}}}$$
$$v=\color{red}{\bf2.04819\times 10^7}\;\rm m/s$$
_____________________________________
b) By the same approach; the charge here is the same as the mentioned example but the mass is doubled.
$$KE =\dfrac{e^2B^2R^2}{2m_{ deuteron}}$$
$$KE=\dfrac{( 1.6\times 10^{-19})^2(1.7)^2(0.25)^2}{2(2\times 1.66\times 10^{-27})}=\bf6.96386\times 10^{-13}\;\rm J $$
$$ KE =\color{red}{\bf 4.352 }\;\rm MeV $$
The velocity is given by
$$KE=\frac{1}{2}m_{\alpha}v^2$$
Thus,
$$v=\sqrt{\dfrac{2KE}{m_\alpha}}=\sqrt{\dfrac{2\times 6.96386\times 10^{-13}}{2\times 1.66\times 10^{-27}}}$$
$$v=\color{red}{\bf 2.04819\times 10^7}\;\rm m/s$$
_____________________________________
c) The frequency is given by
$$f=\dfrac{qB}{2\pi m}$$
Thus,
$$f_\alpha=\dfrac{2eB}{2\pi m_\alpha}=\dfrac{2\times 1.6\times 10^{-19}\times 1.7}{2\pi\times 4\times 1.66\times 10^{-27}}$$
$$f_\alpha=\color{red}{\bf 1.30392\times 10^7}\;Hz$$
$$f_{deuteron}=\dfrac{ eB}{2\pi m_{deuteron}}=\dfrac{ 1.6\times 10^{-19}\times 1.7}{2\pi\times 2\times 1.66\times 10^{-27}}$$
$$f_{deuteron}=\color{red}{\bf 1.30392\times 10^7}\;Hz$$
It is obvious now that both of them are having the same frequency.