Ta có :

a.  $\sqrt{0,36a^{2}}=\sqrt{0,36}.\sqrt{a^{2}}=0,6.\left | a \right |=-0,6a(a<0)$

Vậy $\sqrt{0,36a^{2}}-0,6a(a<0)$

b.  $\sqrt{a^{4}(3-a)^{2}}=\sqrt{a^{4}}.\sqrt{(3-a)^{2}}=\left | a^{2} \right |.\left | 3-a \right|=a^{2}(a-3)(a\geq 3)$

Vậy $\sqrt{a^{4}(3-a)^{2}}=a^{2}(a-3)$

c.  $\sqrt{27.48(1-a)^{2}}=\sqrt{9^{2}.4^{2}.(1-a)^{2}}=\sqrt{9^{2}}.\sqrt{4^{2}}.\sqrt{(1-a)^{2}}(a>1)$

<=> $9.4.\left | 1-a \right |=36(a-1)(a>1<=> 1-a<0)$

Vậy $\sqrt{27.48(1-a)^{2}}=36(a-1)$

d.  $\frac{1}{a-b}\sqrt{a^{4}(a-b)^{2}}=\frac{1}{a-b}.\sqrt{a^{4}}.\sqrt{(a-b)^{2}}$

Vì a > b <=>  $\frac{1}{a-b}.\left | a^{2} \right |.\left | a-b \right |=\frac{1}{a-b}.a^{2}.(a-b)=a^{2}$

Vậy $\frac{1}{a-b}\sqrt{a^{4}(a-b)^{2}}=a^{2}$