Question:

1. Greatest possible diameter
- Let the side of the cube be $a = 7\,\text{cm}$. The hemisphere rests on the top face of the cube. For the hemisphere to fit without overhanging, its diameter $d$ must be at most the side length of the cube.
- Hence the greatest diameter is $$d_{\max}=a = 7\,\text{cm}.$$
- The radius of the hemisphere is therefore $$r = \frac{d_{\max}}{2}=\frac{7}{2}=3.5\,\text{cm}.$$

2. Surface area of the solid
- Curved surface area of the hemisphere: For a hemisphere, the curved surface area (excluding the base circle) is $$\text{CSA}_{\text{hemisphere}} = 2\pi r^{2}.$$
Substituting $r = 3.5\,\text{cm}$,
$$\text{CSA}_{\text{hemisphere}} = 2\pi (3.5)^{2}=2\pi \times 12.25 = 24.5\pi\,\text{cm}^{2}.$$

- Exposed faces of the cube: The cube has six faces, each of area $a^{2}=7^{2}=49\,\text{cm}^{2}$. The top face is completely covered by the hemisphere, so it does not contribute to the external surface. The remaining five faces (four vertical faces + bottom face) are exposed.
$$\text{Area}_{\text{cube}} = 5 \times 49 = 245\,\text{cm}^{2}.$$

- Total surface area:
$$\text{Total SA} = \text{CSA}_{\text{hemisphere}} + \text{Area}_{\text{cube}}
= 24.5\pi + 245\,\text{cm}^{2}$$
or, writing the fractional form,
$$\text{Total SA}=\frac{49\pi}{2}+245\,\text{cm}^{2}.$$

3. Numerical value (optional)
- Using $\pi \approx 3.1416$,
$$\text{Total SA} \approx 245 + 24.5 \times 3.1416 \approx 245 + 76.97 \approx 321.97\,\text{cm}^{2}.$$