Given expression $(\sqrt{x}-\frac{6}{x^{3/2}})^n$. Here, $a = \sqrt{x}$ and $b = -\frac{6}{x^{3/2}}$.

The $r$-th term of the binomial expansion of $(a+b)^n$ is given by

$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.

Substitute $a$ and $b$ in this formula, we get:

$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$.

The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.

This happens when $\frac{n-4r}{2} = 0$, which gives $n = 4r$.

$$ \begin{aligned} & { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\ & (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\ & \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\ & \therefore \alpha=-24 \end{aligned} $$

Now, for coefficient of $x^{-4}$

$$ \begin{aligned} & \frac{n-4 r}{2}=-4 \\\\ & n=4 r-8 \Rightarrow r=3 \\\\ & \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\ & \Rightarrow \lambda=36 \end{aligned} $$