a) YES; b) NO; a) Let the given arithmetic progression is $a_n=a_0+dn,n\in \mathbb{N}$. Consider the sequence $\frac{a_n}{d}=\frac{a_0}{d}+n,n=0,1,\dots$ . It's enough to find to find an infinite geometric progression inside it with $q\in \mathbb{N}$. We reduce $\frac{a_0}{d}$ if possible and could assume $(a_0,d)=1$. It boils down to find $q\in \mathbb{N}$ such that $qa_0\equiv a_0 \pmod d$. To satisfy this, we can choose any $q$ with $q\equiv 1 \pmod d$. $\blacksquare$ b) A stronger claim holds. There exists an infinite arithmetic progression which doesn't contain a finite geometric progression with $3$ or more therms. The following proof is non constructive. The idea is for any $n_1<n_2<n_3$ to rule out the possibility $a_{n_1},a_{n_2},a_{n_3}$ to be a geometric progression. Such condition brings a dependence between $a_0$ and $d$. We exploit the fact all triples $n_1,n_2,n_3\in \mathbb{N}$ are countably many. A quick calculation shows $a_{n_2}^2=a_{n_1}a_{n_3}$ is equivalent to $$(*)\,\,\,\,\,\,\,\, \left(n_1n_3-n_2^2\right)d=a_0\left(2n_2-n_1-n_3 \right)$$It's not possible simultaneously $n_1n_3-n_2^2=0$ and $2n_2-n_1-n_3=0$, hence either $\frac{a_0}{d}=\frac{n_1n_3-n_2^2}{2n_2-n_1-n_3}$ in case $2n_2-n_1-n_3\neq 0$ or $\frac{a_0}{d}=0$ when $2n_2-n_1-n_3= 0$. It means all possible values $\frac{a_0}{d}$ for which $(*)$ holds, when $n_1,n_2,n_3$ vary, are countably many. Since $a_0\in \mathbb{R}$, it follows, for any fixed $d$, we can choose such real $a_0$ for which $(*)$ does not hold for any $n_1,n_2,n_3\in \mathbb{N}$ and the result follows.