Answer
a). $(1)$ increases,
b). $5W: 96dB$
$10W:99dB$
Work Step by Step
a). Since, Intensity level = $10log\frac{I}{I_{0}}=10log\frac{\frac{P}{Area}}{10^{-12}}$
Area is fixed, So, if P increases, intensity level also increases.
b). Intensity level from 5W source= $10log\frac{I}{I_{0}}=10log\frac{\frac{5}{4\times pi\times 10^{2}}}{10^{-12}}=10 log(0.4\times10^{10)}=100-4=96dB$
Intensity level from 10W source= $10log\frac{I}{I_{0}}=10log\frac{\frac{10}{4\times pi\times 10^{2}}}{10^{-12}}=10 log(0.8\times10^{10)}=100-1=99dB$
