(a) The curve \(C\) has polar equation \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \pi\). It is a limacon, symmetric about the line \(\theta = \frac{\pi}{2}\); the completed sketch shows a single loop above the pole.

(b) The line \(l\) has polar equation \[ r\sin\theta = 2 \quad\Rightarrow\quad r = 2\csc\theta \quad (\sin\theta \ne 0). \] Points of intersection satisfy \[ 3 + 2\sin\theta \;=\; 2\csc\theta. \] Multiplying by \(\sin\theta\), \[ 3\sin\theta + 2\sin^2\theta = 2. \] Let \(\sin\theta = s\). Then \[ 2s^2 + 3s - 2 = 0 \quad\Rightarrow\quad s = \frac{-3 \pm 5}{4} \;\Rightarrow\; s = \frac12 \text{ or } s = -2. \] \(\sin\theta = -2\) is impossible, so \(\sin\theta = \frac12\), giving \[ \theta = \frac{\pi}{6},\; \frac{5\pi}{6}. \] At these points, \[ r = 3 + 2\sin\theta = 3 + 2\cdot\frac12 = 4. \] So the points of intersection are \[ \left(4,\; \frac{\pi}{6}\right),\quad \left(4,\; \frac{5\pi}{6}\right). \]

(c) The region \(R\) is enclosed by \(C\) and the line \(l\), and contains the pole. For a given \(\theta\), the boundary closest to the pole is:

• \(r = 3 + 2\sin\theta\) for \(-\pi < \theta \le \frac{\pi}{6}\) and \(\frac{5\pi}{6} \le \theta \le \pi\) (the curve is nearer than the line),
• \(r = 2\csc\theta\) for \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\) (the line is nearer than the curve).

Hence the area of \(R\) is \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi/6} (3 + 2\sin\theta)^2\,d\theta + \frac12 \int_{\pi/6}^{5\pi/6} (2\csc\theta)^2\,d\theta + \frac12 \int_{5\pi/6}^{\pi} (3 + 2\sin\theta)^2\,d\theta. \] It is convenient to rewrite this as \[ \text{Area}(R) = \frac12 \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta \;-\; \frac12 \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta. \]

First, \[ \int_{-\pi}^{\pi} (3 + 2\sin\theta)^2\,d\theta = \int_{-\pi}^{\pi} \bigl(9 + 12\sin\theta + 4\sin^2\theta\bigr)\,d\theta = 22\pi. \]

Next, \[ (3 + 2\sin\theta)^2 - (2\csc\theta)^2 = 9 + 12\sin\theta + 4\sin^2\theta - \frac{4}{\sin^2\theta}, \] so \[ \int_{\pi/6}^{5\pi/6} \Bigl[(3 + 2\sin\theta)^2 - (2\csc\theta)^2\Bigr]\,d\theta = 5\sqrt{3} + \frac{22\pi}{3}. \]

Therefore, \[ \text{Area}(R) = \frac12\left(22\pi - \left(5\sqrt{3} + \frac{22\pi}{3}\right)\right) = \frac12\left(\frac{44\pi}{3} - 5\sqrt{3}\right) = \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}. \]

So the exact area of \(R\) is \[ \boxed{\displaystyle \frac{22}{3}\pi - \frac{5}{2}\sqrt{3}}. \]