Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the total energy of a particle is given by
$$E^2=(mc^2)^2+(pc)^2\tag 1$$
where $p$ the momentum of the proton is given by
$$p=\gamma\; mu$$
where $\gamma=\left[1-\frac{u^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.99^2c^2}{c^2}\right]^{-1/2}=\frac{5}{3}=7.089$
$$p=7.089 (0.99 c)m $$
Plug the known;
$$p=7.089 (0.99 )(3\times 10^8)(1.67\times 10^{-27}) $$
$$p=\color{red}{\bf 3.52\times 10^{-18}}\;\rm kg\cdot m/s$$
Plug into (1),
$$E =\sqrt{m^2c^4+p^2c^2 }$$
$$E =\sqrt{(1.67\times 10^{-27}) ^2(3\times 10^8)^4+(3.52\times 10^{-18})^2(3\times 10^8)^2 }$$
$$E=\color{red}{\bf 1.067\times10^{-9}}\;\rm J$$
$$\color{blue}{\bf [b]}$$
We know that the rest energy of any particle is the same at any reference frame.
So,
$$E'^2=(mc^2)^2+p'^2c^2 $$
Solving for $p'$;
$$p'=\sqrt{\dfrac{E'^2}{c^2}-m^2c^2}$$
Plug the known;
$$p'=\sqrt{\dfrac{(5\times 10^{-10})^2}{(3\times 10^8)^2}-(1.67\times 10^{-27})^2(3\times 10^8)^2}$$
$$p'=\color{red}{\bf 1.59\times 10^{-18}}\;\rm kg\cdot m/s$$
