Answer: (i) The curves intersect when their values of \(r\) are equal:

\(\theta+2=\theta^2\)

So \(\theta^2-\theta-2=0\), hence \((\theta-2)(\theta+1)=0\). Since \(0\leqslant \theta\leqslant \pi\), we take \(\theta=2\).

Then \(r=\theta+2=4\).

The point of intersection is \((4,2)\).

(ii) \(C_1:r=\theta+2\) is an increasing spiral-like curve, starting at \(r=2\) when \(\theta=0\) and reaching \(r=\pi+2\) when \(\theta=\pi\). \(C_2:r=\theta^2\) starts at the origin, increases more slowly at first, and becomes steeper as \(\theta\) increases. The two curves cross at \(\theta=2\).

(iii) The required area is the area between the curves from \(\theta=0\) to \(\theta=2\), since \(C_1\) lies outside \(C_2\) there, together with the line \(\theta=0\). Using the polar area formula,

\(A=\frac12\int_0^2\big((\theta+2)^2-\theta^4\big)\,d\theta\).

Now expand and integrate:

\(A=\frac12\int_0^2(\theta^2+4\theta+4-\theta^4)\,d\theta\)

\(=\frac12\left[\frac{\theta^3}{3}+2\theta^2+4\theta-\frac{\theta^5}{5}\right]_0^2\)

\(=\frac12\left(\frac{8}{3}+8+8-\frac{32}{5}\right)\)

\(=\frac12\left(\frac{232}{15}\right)=\frac{92}{15}.\)

So the area is \(\frac{92}{15}\).

(i) At the point of intersection, the two expressions for \(r\) are equal:

\(\theta+2=\theta^2\).

Rearranging gives \(\theta^2-\theta-2=0\), so

\((\theta-2)(\theta+1)=0\).

Because \(0\leqslant \theta\leqslant \pi\), the valid solution is \(\theta=2\). Substituting back gives

\(r=\theta+2=4\).

Hence the intersection point is \((r,\theta)=(4,2)\).

(ii) For \(C_1\), \(r=\theta+2\): when \(\theta=0\), \(r=2\), and as \(\theta\) increases to \(\pi\), \(r\) increases steadily. So the curve is an outward spiral starting on the positive \(x\)-axis, with radius increasing linearly with \(\theta\).

For \(C_2\), \(r=\theta^2\): when \(\theta=0\), \(r=0\), and the radius increases more slowly at first, then more rapidly as \(\theta\) increases. It also spirals outwards from the origin.

The curves meet at \(\theta=2\), so the sketch should show \(C_2\) inside \(C_1\) for \(0\leqslant \theta\lt 2\), then crossing at that angle.

(iii) The required region is enclosed by \(C_1\), \(C_2\), and the line \(\theta=0\). Since the curves meet at \(\theta=2\), the area is

\(A=\frac12\int_0^2\left((\theta+2)^2-\theta^4\right)\,d\theta\).

Expand the integrand:

\(A=\frac12\int_0^2\left(\theta^2+4\theta+4-\theta^4\right)\,d\theta\).

Integrating term by term gives

\(A=\frac12\left[\frac{\theta^3}{3}+2\theta^2+4\theta-\frac{\theta^5}{5}\right]_0^2\).

Substitute \(\theta=2\):

\(A=\frac12\left(\frac{8}{3}+8+8-\frac{32}{5}\right)\).

Putting this over a common denominator:

\(\frac{8}{3}+16-\frac{32}{5}=\frac{40}{15}+\frac{240}{15}-\frac{96}{15}=\frac{184}{15}\), so

\(A=\frac12\cdot\frac{184}{15}=\frac{92}{15}.\)

Therefore the area bounded by the two curves and the line \(\theta=0\) is \(\frac{92}{15}\).