$textbf{Find:}$ The number of half-lives required to reduce the sample to $0.125 N_o$
$N_o$ = initial amount, $N_{final} = 0.125 N_o$ , and $n$ is the number of half-lives.

To solve for $n$, we obtain the following formula from the decay equation:

$$
begin{align*}
N &= dfrac{N_o}{2^n} \
0.125 N_o &= dfrac{N_o}{2^n} \
2^n &= 8 \
end{align*}
$$

Getting the log of base 2 on both sides of the equation we have:

$$
begin{align*}
log_{2}(2^n) &= log_{2}8 \
n &= 3
end{align*}
$$

Hence, it will require $boxed{text{3 half-lives}}$ to reduce the sample to $dfrac{1}{8}$ of its original amount.

$$
textbf{3 half-lives}
$$

The observed activity from the fossil sample is $0.190$ Bq, whereas the half-life of Carbon is $0.116$ Bq. Therefore, we can expect that the age of the fossil is $boxed{text{less than $5730$ years old}}$

The sample is less than $5730$ years old

The physical idea is mass-energy equivalence. When nucleons are bond together part of their mass is used for bonding energy. If the mass of the products is lower than the mass of the reactants then the huge amount of energy is released since $E=Delta m c^2$ and $c$ is really big.

When the first half-life passes half of the original number of nuclei decay. When another one passes half of the remainder decays which is a quarter of the starting number so the correct answer is $N/4$.

a) Given the reaction, we can solve for the mystery element by calculating Atomic mass (A) and Atomic number (Z), such that:

Solving for A, we have:

236 = 137 + 4 + A

A = 95

Solving for Z:

92 = 55 + Z
Z = 37

The mystery nucleus here is $_{37}^{95}$X, using this as our reference in the periodic table we have: $boxed{_{37}^{95} Rb}$

b) Given:

$Delta$ m = 0.205 u

To calculate for the energy released we use the formula:

$$
E = Delta m times E_u
$$

where $E_u$ = $931.5$ MeV

$$
begin{align*}
E &= (0.205 text{u}) times (931.5 text{MeV}) \
&= 190.026 text{MeV}
end{align*}
$$

Hence, the system released $boxed{190 text{ MeV}}$

a) $_{37}^{95} Rb$

b) $190$ MeV

Given:

half-life = $3.8$ d

$R_{final} = 0.25R_{initial}$

Formula:

time = $dfrac{text{half-life}}{0.693} ln left (dfrac{R_{initial}}{R_{final}} right )$

We can calculate the time using the formula above, such that:

$$
begin{align*}
time &= dfrac{3.8 text{d}}{0.693} ln left (dfrac{R_{initial}}{0.25 R_{initial}} right ) \
&= dfrac{3.8 text{d}}{0.693} ln (dfrac{1}{0.25}) \
&= 7.6 d
end{align*}
$$

Hence, the time it takes for the radon atoms to decrease to 0.25 times its original amount is $boxed{text{7.6 d}}$