$(a)$ Electric field due to given charge distribution is

$E=-\frac{d V}{d r}=-\frac{d}{d r} \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$

$=\frac{-q}{4 \pi \varepsilon_{0}} \frac{d}{d r}\left(\frac{e^{-\alpha r}}{r}\right)$

$=\frac{-q}{4 \pi \varepsilon_{0}}\left(\frac{-\alpha r e^{-\alpha \alpha_{r}}-e^{-\alpha r}}{r^{2}}\right)$

$=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha r} \cdot\left(\frac{\alpha r+1}{r^{2}}\right)$

Elcctric field at $r=\frac{1}{1}$ is

Electric field at $r=\frac{1}{\alpha}$ is

$E\left(r=\frac{1}{\alpha}\right)=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha \times \frac{1}{\alpha}}\left(\frac{\alpha \times \frac{1}{\alpha}+1}{1 / \alpha^{2}}\right)$

$=\frac{(q / e)}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2}$

Flux through a sphere of radius $\frac{1}{\alpha}$ is

$\phi=\int E \cdot d A$

For spherical distribution,

$E. d A =E d A \cos 0^{\circ}=E d A$

and $E=$ uniform

So, we have

$\therefore \quad \phi=E \int d A$

$=\frac{ q / e}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2} \cdot 4 \pi\left(\frac{1}{\alpha}\right)^{2}$

or $\quad \phi=\frac{2 q}{\varepsilon_{0} e}$

From Gauss' law, we have $\phi=\frac{q_{\text {enclosed }}}{\varepsilon_{0}}$

So, charge enclosed in given sphere is $\frac{2 q}{e}$.