$$\begin{array}{ll} \text { Let } & x \in A \cup B \\ \Rightarrow & x \in A \text { and } x \in B \\ \Rightarrow & x \in C \text { and } x \in C \quad [\because A \subset C \text { and } B \subset C]\\ \Rightarrow & x \in C \Rightarrow A \cup B \subset C \end{array}$$

Hence, given statement is true.

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

$\left[\because A-B=A \cap B^{\prime}\right]$

$\begin{array}{lr}=(A \cup B) \cap\left(A \cup A^{\prime}\right)=(A \cup B) \cap U & {\left[\because A \cup A^{\prime}=U\right]} \\ =A \cup B=R H S & {[\because A \cap U=A]}\end{array}$

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

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

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

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

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