### (a) Using similarity of $\triangle$AHK and $\triangle$ABC

**Step 1 — Write the correspondence.**
$\triangle AHK \sim \triangle ABC$ means $A \leftrightarrow A,\ H \leftrightarrow B,\ K \leftrightarrow C$.

**Step 2 — Write the ratio of corresponding sides.**
$$\frac{AH}{AB} = \frac{HK}{BC} = \frac{AK}{AC}$$

**Step 3 — Use the two known ratios.**
$$\frac{HK}{BC} = \frac{AK}{AC} \ \Rightarrow\ \frac{7}{3{\cdot}5} = \frac{10}{AC}$$

**Step 4 — Solve for AC.**
$$AC = \frac{10 \times 3{\cdot}5}{7} = \frac{35}{7} = 5 \text{ cm}$$

$$\boxed{AC = 5 \text{ cm}}$$

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### (b) Using Basic Proportionality Theorem

**Step 1 — Convert the given ratio.**
$$\frac{PQ}{XQ} = \frac{7}{3} \ \Rightarrow\ \frac{PX + XQ}{XQ} = \frac{7}{3}$$

**Step 2 — Find PX : XQ.**
$$\frac{PX}{XQ} + 1 = \frac{7}{3} \ \Rightarrow\ \frac{PX}{XQ} = \frac{7}{3} - 1 = \frac{4}{3}$$

**Step 3 — Apply BPT (Thales theorem), since XY $\parallel$ QR.**
$$\frac{PX}{XQ} = \frac{PY}{YR} = \frac{4}{3}$$

**Step 4 — Use PR = PY + YR = 6·3 cm.**
$$YR = \frac{3}{4+3} \times PR = \frac{3}{7} \times 6{\cdot}3$$

**Step 5 — Calculate.**
$$YR = \frac{18{\cdot}9}{7} = 2{\cdot}7 \text{ cm}$$

$$\boxed{YR = 2{\cdot}7 \text{ cm}}$$