$$\begin{aligned} \mathrm{LHS} & =A-(A-B)=A-\left(A \cap B^{\prime}\right) \quad \left[\because A-B=A \cap B^{\prime}\right]\\ & =A \cap\left(A \cap B^{\prime}\right)^{\prime}=A \cap\left[A^{\prime} \cup\left(B^{\prime}\right)^{\prime}\right] \quad \left[\because(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\right]\\ & =A \cap\left(A^{\prime} \cup B\right) \quad \left[\because\left(A^{\prime}\right)^{\prime}=A\right]\\ & =\left(A \cap A^{\prime}\right) \cup(A \cap B)=\phi \cup(A \cap B) \\ & =A \cap B=R H S \end{aligned}$$

$\mathrm{LHS}=A-(A \cap B)=A \cap(A \cap B)^{\prime} \quad\left[\because A-B=A \cap B^{\prime}\right]$

$=A \cap\left(A^{\prime} \cup B^{\prime}\right) \quad\left[\because(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\right]$

$$ \begin{aligned} & =\left(A \cap A^{\prime}\right) \cup\left(A \cap B^{\prime}\right)=\phi \cup\left(A \cap B^{\prime}\right) \\ & =A \cap B^{\prime} \quad [\because \phi \cup A=A]\\ & =A-B=\mathrm{RHS} \end{aligned}$$

$\begin{aligned} \mathrm{LHS} & =(A \cup B)-B=(A \cup B) \cap B^{\prime} \quad \left[\because A-B=A \cap B^{\prime}\right]\\ & =\left(A \cap B^{\prime}\right) \cup\left(B \cap B^{\prime}\right)=\left(A \cap B^{\prime}\right) \cup \phi \quad \left[\because B \cap B^{\prime}=\phi\right]\\ & =A \cap B^{\prime} \quad [\because A \cup \phi=A]\\ & =A-B=\mathrm{RHS}\end{aligned}$

Since, $$T=\left\{x \left\lvert\, \frac{x+5}{x-7}-5=\frac{4 x-40}{13-x}\right.\right\}$$

$\begin{array}{ll}\because & \frac{x+5}{x-7}-5=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{x+5-5(x-7)}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{x+5-5 x+35}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{-4 x+40}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & -(4 x-40)(13-x)=(4 x-40)(x-7)\end{array}$

$\Rightarrow \quad(4 x-40)(x-7)+(4 x-40)(13-x)=0$

$$\begin{array}{rlr} \Rightarrow & (4 x-40)(x-7+13-x) =0 \\ \Rightarrow & 4(x-10) 6 =0 \\ \Rightarrow & 24(x-10) =0 \\ \Rightarrow & x =10 \\ \therefore & T =\{10\} \end{array}$$

Hence, T is not an empty set.

$$\begin{array}{ll} \text { Let } & x \in A \cap(B \cup C) \\ \Rightarrow & x \in A \text { and } x \in(B \cup C) \\ \Rightarrow & x \in A \text { and }(x \in B \text { or } x \in C) \\ \Rightarrow & (x \in A \text { and } x \in B) \text { or }(x \in A \text { and } x \in C) \\ \Rightarrow & x \in A \cap B \text { or } x \in A \cap C \\ & x \in(A \cap B) \cup(A \cap C) \\ \Rightarrow & A \cap(B \cup C) \subset(A \cap B) \cup(A \cap C)\quad \text{... (i)} \end{array}$$

$\begin{array}{ll}\text { Again, let } & y \in(A \cap B) \cup(A \cap C) \\ \Rightarrow & y \in(A \cap B) \text { or } y \in(A \cap C) \\ \Rightarrow & (y \in A \text { and } y \in B) \text { or }(y \in A \text { and } y \in C) \\ \Rightarrow & y \in A \text { and }(y \in B \text { or } y \in C) \\ \Rightarrow & y \in A \text { and } y \in B \cup C \\ \Rightarrow & y \in A \cap(B \cup C) \\ \Rightarrow & (A \cap B) \cup(A \cap C) \subset A \cap(B \cup C)\quad \text{.... (ii)}\end{array}$

From Eqs.(i) and( ii),

$$A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$$