1. **Stating the problem:** Given $\log_3 5 = a$ and $\log_5 7 = b$, find the values of $\log_{15} 49$ and $\log_{49} \sqrt{15}$.\n\n2. **Recall the change of base formula:** For any positive numbers $x, y, c$ with $x \neq 1$ and $y \neq 1$,\n$$\log_x y = \frac{\log_c y}{\log_c x}.$$\nWe will use base 3 for convenience since $a$ is given in base 3.\n\n3. **Express given logs in terms of $a$ and $b$: **\n- $\log_3 5 = a$ (given)\n- $\log_5 7 = b$ (given)\nUsing change of base, $\log_3 7 = \log_3 5 \times \log_5 7 = a \times b$.\n\n4. **Calculate $\log_{15} 49$: **\nWrite 15 and 49 in terms of prime factors:\n- $15 = 3 \times 5$\n- $49 = 7^2$\nUsing change of base to base 3,\n$$\log_{15} 49 = \frac{\log_3 49}{\log_3 15} = \frac{\log_3 7^2}{\log_3 (3 \times 5)} = \frac{2 \log_3 7}{\log_3 3 + \log_3 5} = \frac{2 (a b)}{1 + a}.$$\n\n5. **Calculate $\log_{49} \sqrt{15}$: **\nWrite $\sqrt{15}$ as $15^{1/2}$ and 49 as $7^2$,\n$$\log_{49} \sqrt{15} = \frac{\log_3 \sqrt{15}}{\log_3 49} = \frac{\log_3 15^{1/2}}{\log_3 7^2} = \frac{\frac{1}{2} \log_3 15}{2 \log_3 7} = \frac{\frac{1}{2} (\log_3 3 + \log_3 5)}{2 (\log_3 7)} = \frac{\frac{1}{2} (1 + a)}{2 (a b)} = \frac{1 + a}{4 a b}.$$\n\n**Final answers:**\n$$\log_{15} 49 = \frac{2 a b}{1 + a}$$\n$$\log_{49} \sqrt{15} = \frac{1 + a}{4 a b}$$