Given that, change in emitter current, $∆I_E=8 \mathrm{mA}$ and change in collector current, $∆I_C=7.9 \mathrm{mA}$.
We know that,
$\mathrm{\alpha }=\frac{{∆I}_C}{{∆I}_E}$
$\Rightarrow \alpha =\frac{7.9}{8}$
$\Rightarrow \mathrm{\alpha }\approx 0.99$
Also, we know that
$\mathrm{\beta }=\frac{\alpha }{1-\alpha }$
$\Rightarrow \beta =\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}=\frac{7.9}{8-7.9}$
$\Rightarrow \mathrm{\beta }=\frac{7.9}{0.1}=79$
Hence, the required answer is $\mathrm{\alpha }=0.99$ and $\mathrm{\beta }=79$.