If $f(x)=\frac{2}{3} a^2 x^3-\frac{5 a}{2} x^2+3 x+b$, then the set of values of b for which local extrema of the function f(x) are positive and maximum occurs at $x=\frac{1}{3}$, is

$(-3 / 8, \infty)$

We have,

$f(x) =\frac{2}{3} a^2 x^3-\frac{5 a}{2} x^2+3 x+b$            .........(i)

$\Rightarrow f'(x) =2 a^2 x^2-5 a x+3$ and $f''(x)=4 a^2 x-5 a$

It is given that f(x) attains maximum at $x=1 / 3$. Therefore,

$f'\left(\frac{1}{3}\right)=0$ and $f''\left(\frac{1}{3}\right)<0$

$\Rightarrow \frac{2 a^2}{9}-\frac{5 a}{3}+3=0$ and $\frac{4 a^2}{3}-5 a<0$

$\Rightarrow 2 a^2-15 a+27=0$ and $4 a^2-15 a<0 $

$\Rightarrow (2 a-9)(a-3)=0$ and $a(4 a-15)<0$

$\Rightarrow a=3, \frac{9}{2}$ and $0<a<\frac{15}{4} \Rightarrow a=3$

∴  $f'(x)=18 x^2-15 x+3$ and $f''(x)=36 x-15$

At extreme points, we have

$f'(x)=0 \Rightarrow 6 x^2-5 x+1=0 \Rightarrow x=\frac{1}{2}, \frac{1}{3}$

Clearly, $f''\left(\frac{1}{2}\right)>0$ and $f''\left(\frac{1}{3}\right)<0$

So, $x=\frac{1}{2}$ is the point of local minimum and $x=\frac{1}{3}$ is the point of local maximum.

Putting a = 3 in (i), we get

$f(x)=6 x^3-\frac{15}{2} x^2+3 x+b$

The extreme values of f(x) are

$f\left(\frac{1}{2}\right)=\frac{6}{8}-\frac{15}{8}+\frac{3}{2}+b=\frac{3}{8}+b$

and, $f\left(\frac{1}{3}\right)=\frac{6}{27}-\frac{15}{18}+1+b=\frac{7}{18}+b$

For these extreme values to be positive, we must have

$\frac{3}{8}+b>0$ and $\frac{7}{18}+b>0$

$\Rightarrow b>-\frac{3}{8}$ and $b>-\frac{7}{18} \Rightarrow b>-\frac{3}{8} \Rightarrow b \in\left(-\frac{3}{8}, \infty\right)$