We know that $${p \leftrightarrow q}$$  is true if $$p$$ and $$q$$ both are true or false.
so, $${p \leftrightarrow \, \sim q}$$   is true when if $$p$$ and $$\sim q$$  is true.
i.e., $$p$$ is true and $$q$$ is false.
or $$p$$ and $$\sim q$$  is false, i.e. $$p$$ is false and $$q$$ is true.

$${ \sim p \vee q} :$$   Raju is not tall or he is intelligent.

The truth value of $$ \sim \left( { \sim p} \right) \leftrightarrow p\,$$   as follow

$$p$$	$$ \sim p$$	$$ \sim \left( { \sim p} \right)$$	$$ \sim \left( { \sim p} \right) \to p$$	$$p \to \sim \left( { \sim p} \right)$$	$$ \sim \left( { \sim p} \right) \leftrightarrow p$$
T	F	T	T	T	T
F	T	F	T	T	T

Since last column of above truth table contains only $$T.$$
Hence, $$ \sim \left( { \sim p} \right) \to p\,$$   is a tautology.

The inverse of the given statement is ‘If $$x$$ is not zero then we devide by $$x$$’

We have

$$p$$	$$q$$	$$ \sim p$$	$$p \to q$$	$$ \sim p \to q$$	$$\left( { \sim p \to q} \right) \to q$$	$$\left( {p \to q} \right) \to \left( {\left( { \sim p \to q} \right) \to q} \right)$$
T	F	F	F	T	F	T
T	T	F	T	T	T	T
F	F	T	T	F	T	T
F	T	T	T	T	T	T

∴ It is tautology.

Last column shows that result is neither a tautology nor a contradiction.

Let $$p$$ : I become a teacher.
$$q$$ : I will open a school
Negation of $$p \to q$$  is $$ \sim \left( {p \to q} \right) = {p^ \wedge } \sim q$$
i.e. I will become a teacher and I will not open a school.

We know that the contropositive of $$p \to q$$   is $$ \sim q \to \, \sim p.$$   So contra positive of $$p \to \left( { \sim q \to \, \sim r} \right)$$    is
$$\eqalign{ & \sim \left( { \sim q \to \, \sim r} \right) \to \, \sim p \equiv \,\, \sim q \wedge \left[ { \sim \left( { \sim r} \right)} \right] \sim p \cr & \because \,\, \sim \left( {p \to q} \right) \equiv p \, \wedge \sim q \equiv \,\, \sim q \wedge r \to \, \sim p \cr} $$

Since, $$ \sim \left( {p \vee q} \right) \equiv \, \sim p \, \wedge \sim q\,\,$$     (By De-Morgans' law)
$$\therefore \,\, \sim \left( {p \vee q} \right) \ne \,\, \sim p \, \vee \sim q$$
$$\therefore \left( D \right)$$  is the false statement.

$${ \sim p} :$$  Ashok does not work hard
Use $${' \to '}$$  symbol for then
$$\left( { \sim p \to q} \right)$$   mean = If Ashok does not work hard then he gets good grade.