\(a * a^{-1} = e = -1\)

\(\therefore a^{-1} = -1\)

\(180 = \pi rads\)

\(1 = \frac{\pi}{180}\)

\(\therefore 75 = \frac{\pi}{180} \times 75 \)

= \(\frac{5\pi}{12}\)

\(\log_{9} 3 = \log_{9} (9^{\frac{1}{2}}) = \frac{1}{2}\log_{9} 9 = \frac{1}{2}\)

\(\frac{1}{2} + 2x = 1 \implies 2x = \frac{1}{2}\)

\(x = \frac{1}{4}\)

\(\cos (x + y) = \cos x \cos y - \sin x \sin y\)

\(\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\)

= \((0 \times \frac{1}{2}) - (1 \times \frac{\sqrt{3}}{2})\)

= \(0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\)

Using remainder theorem, the remainder when \(5x^{3} + 2x^{2} - 7x -5\) is divided by (x - 2) = f(2)

\(f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5\)

= 29

\(f(x) = \frac{3x + 1}{x^{2} - 1}\)

\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)

\(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}} \)

\(\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}\)

= \(\frac{2}{3} - \frac{3}{2}\)

= \(\frac{-5}{6}\)

\(3x^{2} + 4x + 1 > 0 \)

\(3x^{2} + 3x + x + 1 > 0\)

\(3x(x + 1) + 1(x + 1) > 0\)

\((3x + 1)(x + 1) > 0\)

\(3x + 1 > 0 \implies 3x > -1 \)

\(x > -\frac{1}{3}\)

\(x + 1 > 0 \implies x > -1\)

\(x > -1, x > -\frac{1}{3}\)

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1, we get that

\(-2a = 2\implies a = -1\)

\(2b = 4\implies b = 2\)

\(r^{2} - a^{2} - b^{2} = -2\)

\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2\implies r^{2} = -2+ 1 + 4 = 3\)

\(r = \sqrt{3}\)

\(f(x) = p + qx\)

\(f(1) = p + q(1) \implies p + q = 7 .... (1)\)

\(f(5) = p + 5q = 19 .....(2)\)

Solving for p and q using simultaneous equation, p = 4, q = 3

\(f(3) = 4 + 3(3) = 13\)