### (i) Diagonal of the cube (edge 6 cm)

**Step 1 — Recall the formula for the body diagonal of a cube.**
$$d = a\sqrt3$$

**Step 2 — Substitute $a = 6$ cm.**
$$d = 6\sqrt3 \text{ cm} \approx 10{\cdot}39 \text{ cm}$$

$$\boxed{d = 6\sqrt3 \text{ cm}}$$

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### (ii) Volume of the cube (edge 7 cm)

**Step 3 — Recall the formula.**
$$V = a^3$$

**Step 4 — Substitute $a = 7$ cm.**
$$V = 7^3 = 343 \text{ cm}^3$$

$$\boxed{V = 343 \text{ cm}^3}$$

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### (iii) (a) Curved surface area of the hemisphere (r = 7 cm)

**Step 5 — Recall the formula.**
$$\text{CSA of hemisphere} = 2\pi r^2$$

**Step 6 — Substitute $r = 7$ cm and $\pi = \dfrac{22}{7}$.**
$$= 2 \times \frac{22}{7} \times 7 \times 7 = 2 \times 22 \times 7$$

**Step 7 — Calculate.**
$$= 308 \text{ cm}^2$$

$$\boxed{\text{CSA} = 308 \text{ cm}^2}$$

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### (iii) (b) Surface area of the cuboid from two 4 cm cubes

**Step 8 — Find the dimensions of the resulting cuboid.**
Joining two cubes of edge 4 cm end-to-end :
$$l = 4 + 4 = 8 \text{ cm}, \qquad b = 4 \text{ cm}, \qquad h = 4 \text{ cm}$$

**Step 9 — Apply the TSA formula.**
$$\text{TSA} = 2(lb + bh + hl)$$

**Step 10 — Substitute.**
$$= 2(8\times4 + 4\times4 + 4\times8) = 2(32 + 16 + 32)$$

**Step 11 — Calculate.**
$$= 2 \times 80 = 160 \text{ cm}^2$$

$$\boxed{\text{Surface area} = 160 \text{ cm}^2}$$