Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Tritium ($ ^3 \rm{H} $) is a radioactive isotope of hydrogen, with an atomic number $ Z = 1 $ and a neutron count $ N = 2 $.
This means that Tritium undergoes beta-minus ($ \beta^- $) decay, as seen in Appendix C.
In this type of decay, a neutron in the nucleus transforms into a proton, emitting an electron (beta particle) and an antineutrino.
Hence, tritium decays into the stable isotope of helium, $ ^3 \rm{He} $, through the reaction of
$$
^3 \rm{H} \to \;^3\rm{He} + \beta^-
$$
Thus, the daughter nucleus of tritium decay is $\boxed{ ^3 \rm{He}} $.
$$\color{blue}{\bf [b]}$$
The half-life ($ t_{1/2} $) of tritium is given as 12.33 years.
Recalling that the decay rate, $ r $, can be calculated from the relationship:
$$
r =\dfrac{1}{\tau}= \frac{\ln(2)}{t_{1/2}}
$$
from $ N = N_0 \left( \frac{1}{2} \right)^{t / t_{1/2}} $; Substituting $ t_{1/2} = 12.33$ years,
$$
r = \frac{\ln(2)}{12.33 \; \rm{years}} \times \frac{1 \; \rm{year}}{3.156 \times 10^7 \; \rm{s}} = \color{red}{\bf1.78 \times 10^{-9}} \; \rm{s}^{-1}
$$
Therefore, the lifetime ($ \tau $) of tritium is given by
$$
\tau = \frac{1}{r} = \frac{1}{1.83 \times 10^{-9} \; \rm{s}^{-1}} =\color{red}{\bf 5.614 \times 10^8 }\; \rm{s}
$$
