$A$ radioactive nucleus will not emit

The penetrating power of an $\alpha$-particle is:

In the following nuclear reaction,identify $x$: $_{29}Cu^{64} \to _{28}Ni^{64} + x$

In the decay sequence:
$^{238}_{92}U$ $\xrightarrow{- x_1} ^{234}_{90}Th$ $\xrightarrow{- x_2} ^{234}_{91}Pa$ $\xrightarrow{- x_3} ^{234}_{92}Z$ $\xrightarrow{- x_4} ^{230}_{90}Th$
$x_1, x_2, x_3$ and $x_4$ are particles/radiation emitted by the respective isotopes. The correct option$(s)$ is/are-
$(1)$ $Z$ is an isotope of uranium
$(2)$ $x_2$ is $\beta^{-}$
$(3)$ $x_1$ will deflect towards negatively charged plate
$(4)$ $x_3$ is $\gamma$-ray

$A$ radioactive nuclide $X$ decays at the rate of $1.00 \times 10^5 \ disintegration \ s^{-1} \ g^{-1}$. Radium decays at the rate of $3.70 \times 10^{10} \ disintegration \ s^{-1} \ g^{-1}$. The activity of $X$ in millicuries $g^{-1}$ $(mCi \ g^{-1})$ is

$A$ radioisotope will not emit