Let e be the identity element in Q.
Then for every a € Q such that
A*e = e * a = a
Given \( a^{*} \mathrm{~b} =\frac{a b}{8} \)
Now, \(a^{*} \mathrm{e} =a \)
\(\frac{a e}{8} =a \)
\(\mathrm{e} =8 \in \mathrm{Q} \)
Hence 8 is the identity element for given operation.

The last columns contains only 'T'
\(\therefore\) The given statement is a tautology. 

\(N \times N \rightarrow N\) given by (a, b) \(\Rightarrow\) a - b is not binary operation, because as the image of (2, 5) under '-' is 2 - 5 \(=-3 \notin N\)
Similarly, \(\div\) is not abinary operation on N.
Forexample \(1 * 2=1 \div 2=\frac{1}{2} \notin N\)

\( \mathrm{a} * \mathrm{~b} =\frac{a b}{5} \)
\(2 *(x * 5) =10 \)
\(2 *\left(\frac{5 x}{5}\right) =10 \)
\(2 * x =10 \)
\(\frac{2 x}{5} =10 \Rightarrow x=\frac{50}{2}=25 \)

Clearly 1 is the identity element of G
w² . a^-1 = e ⇒ w² . a^-1 = a^-1 = w
Since w² . w = w³ = 1
Inverse of w² is w.

Let a \(\in \) S and e be the identity element.
Then a * e = a ⇒ \(\frac { ae }{ 2 } \) = a
⇒ ae = 2a ⇒ e = 2
Let a^-1 be the inverse of \(\frac { 1 }{ 2 } \)
Then \(\frac { 1 }{ 2 } \) * a^-1 = e = \(\frac { 1 }{ 2 } \) * a^-1 = e
⇒ \(\frac { { a }^{ -1 } }{ 2 } \frac { { a }^{ -1 } }{ 2 } \) = 2 ⇒ a^-1 = 8.

Let a be any element and e be the identity element.
The a * e = e * a = a
a * e = a ⇒ a + e -1 = a ⇒ e-1 = 0 ⇒ e = 1
∴ The identity element is 1

∴p v (q Λ r) is a contingency

The last column contains only T
The given statement is a tautology