Finding half-life of a nuclide based on gas release
For this we can first write the equation for the reaction again:
$$MH_2 =&gt; He_{(g)} + Noble gas + H_{2(g)}$$
Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then we can use the first order integrated rate law now:
$$ln(N_f) = -kt + ln(N_i)$$
$$ln(0.025-0.015) = -k(80 mins) + ln(0.025)$$
$$k = 0.0115 min^{-1}$$
Now that we have k, we can solve for the half-life.
$$t_{1/2}=\frac{ln(2)}{k}$$
$$t_{1/2}=\frac{ln(2)}{0.0115min^{-1}} \approx 61 mins$$