Note that $an+b$ is a perfect $k{\rm th}$ power iff there is a $z$ such that $z^k\equiv b\pmod{a}$. (a) Take $a=71$, $g$ to be a primitive root modulo $a$ and $b\equiv g^{10}\pmod{71}$. Clearly taking $X\equiv g^5\pmod{71}$ ensures $X^2\equiv b\pmod{71}$ and taking $Y\equiv g^2\pmod{71}$ ensures $Y^5\equiv b\pmod{71}$. Suppose there is a $z$ for which $z^7\equiv b\pmod{71}$. Let $z\equiv g^i$. Then, $g^{7i}\equiv g^{10}\pmod{71}$. This yields $7i\equiv 10\pmod{70}$. This is a contradiction with mod 7. (b) I do the proof for the case $a$ is odd, the case $a$ even should be similar. The answer is yes. Let $a=\textstyle \prod_{1\le i\le L}p_i^{e_i}$ where $p_i$ are distinct odd primes. Let $b\equiv g_i^{v_i}\pmod{p_i^{e_i}}$, $X\equiv g_i^{k_i}$ and $Y\equiv g_i^{\ell_i}$, where $g_i$ is a primitive root, and $X^5\equiv Y^2\equiv b\pmod{p_i^{e_i}}$. We have $g_i^{5k_i}\equiv g_i^{2\ell_i}\pmod{p_i^{e_i}}$, so $5k_i\equiv 2\ell_i\pmod{\varphi(p_i^{e_i})}$, using Euler's theorem. Since $\varphi(p_i^{e_i})$ is even, we must have $2\mid k_i$, set $k_i=2u_i$. Then, $b\equiv g_i^{10u_i}\pmod{p_i}$. Now take $A\equiv g_i^{u_i}\pmod{p_i^{e_i}},\forall i\in\{1,\dots,L\}$---such $A$ exists by the Chinese Remainder Theorem. Then, $A^{10}\equiv b\pmod{n}$, completing the proof.