If the number of integral terms in the expansion of ${\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)}^n$ is exactly $33$, then the least value of $n$ is

${\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)}^n$

General term $T_{r+1}={}^{n}C_r{\left(3^{\frac{1}{2}}\right)}^{n-r}{\left(5^{\frac{1}{8}}\right)}^r$

$T_{r+1}={}^{n}C_r{3}^{\frac{n-r}{2}}{5}^{\frac{r}{8}}$.

$\frac{n-r}{2}, \frac{r}{8}$ must be integer.

$\frac{n-r}{2}\in I\Rightarrow n-r=0, 2, 4, 6, 8.......$

$\frac{r}{8}\in I\Rightarrow r=0, 8, 16, 24, 32....$

For rational term common powers$=0, 8, 16, 24.......$

Number of integral term$=33$

$T_s=a+\left(s-1\right)d \Rightarrow T_{33}=0+\left(33-1\right)8=256$.

Value of $n=256$.

We can cross check to verify answer.