Answer
(a) The intensity increases by 125%.
(b) The intensity level increases by $3.52~dB$
Work Step by Step
(a) If the original amplitude was $A$, then the new amplitude is $1.5~A$. Since the intensity is proportional to the square of the amplitude, the new intensity is proportional to $(1.5~A)^2 = 2.25~A^2$, while the original amplitude was proportional to $A^2$.
Let $I_1$ be the original intensity and let $I_2$ be the new intensity. Note that $I_2 = 2.25~I_1$. We can see that the intensity increases by 125%.
(b) We can find the new intensity level $\beta_2$:
$\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$
$\beta_2-\beta_1 = 10~log~\frac{I_2}{I_1}$
$\beta_2-\beta_1 = 10~log~\frac{2.25~I_1}{I_1}$
$\beta_2-\beta_1 = 10~log~2.25$
$\beta_2 = \beta_1 + 10~log~2.25$
$\beta_2 = \beta_1+3.52~dB$
The intensity level increases by $3.52~dB$
