b. suppose there exist an infinite AP with this property. suppose it is $a,a+b,a+2b,....$. suppose $(a,b)=1$ , then by Dirichlet theorem , there exist infinitely many terms of this AP whose magnitude is a positive prime number . hence a contradiction ! now consider the case $(a,b)=g>1$ then suppose $a=gx,b=gy$ . again $x,x+y,...$ contains infinitely many primes and after a certain extent , all such primes are strictly greater in magnitude than the highest prime factor of $g$, then again they can;t be perfect powers.a contradiction ! hence no such infinite AP exists.

Its a result which states that for any natural number $n$, there exists an increasing AP with each term a perfect power($>1$) of an integer. See, 250 problems in NT by Sierpienski for example.

Remark It did seems that part $a)$ is indeed a problem in Sierpinski, and he himself put a solution by Polish number theorist Andrzej Schinzel. I did not have a time to read it, and maybe, I can post it some other time. Below is my (messy) solution. $a)$ Yes, and in fact, one can construct arbitrarily (but finite) long arithmetic sequences, where, each term is a perfect power. The idea is to proceed as follows. For the sake of simplicity, suppose, we have three perfect powers, $x_1=a_1^{n_1}$, $x_2=a_2^{n_2}$, and $x_3=a_3^{n_3}$, that are in arithmetic progression. Our idea is to generate, somehow, a fourth number, which will be in an arithmetic progression; as well as, a perfect power ($>1$) by itself. Let, $T=\Theta^{n_1n_2n_3}$, for some $\Theta$ to be determined later. We first claim that $Tx_1$, $Tx_2$ and $Tx_3$ are all perfect powers. To see this, note that, $$ T x_1 = (\Theta^{n_2n_3}a_1)^{n_1}, \, T x_2 = (\Theta^{n_1n_3}a_2)^{n_2}, \, \text{ and, } T x_3 = (\Theta^{n_1n_2}a_3)^{n_3}. $$Now, no matter what $\Theta$ is, first three terms of the sequence $\tilde{x}_i = Tx_i$ are all perfect powers. Next, the common difference of this new arithmetic sequence is, simply, $d=\Theta^{n_1n_2n_3}(a_3^{n_3}-a_2^{n_2})$. Thus, $$ \tilde{x}_4 = d+ \tilde{x}_3=\Theta^{n_1n_2n_3}(2a_3^{n_3}-a_2^{n_2}). $$Let, $\Theta=(2a_3^{n_3}-a_2^{n_2})^{\gamma}$, for some $\gamma>1$. With this choice of $\Theta$, we have, $$ \tilde{x}_4 =(2a_3^{n_3}-a_2^{n_2})^{\gamma n_1 n_2 n_3+1}=a_4^{n_4}, $$for some $a_4$, and $n_4$, and moreover, $n_4$ is coprime with the rest of three numbers. As an execution of the argument, start with two terms, for instance, $a_1=2$, $n_1=2$, $a_2=3$, and $n_2=3$. With the notation of above, we have, for, $\Theta=50$, $\gamma=1$, $T=50^6$, we have, $$ Tx_1 = (2 \cdot 50^3)^2, Tx_2 = (3 \cdot 50^2)^3, $$and $\tilde{x}_3$ is computed by, $$ \tilde{x}_3 = 2Tx_2 - Tx_1 = 50^7. $$Continuing in this manner, we can construct an arithmetic sequence with length $2003$, each term of whom is a perfect power. Remark This approach is rather quick-n-dirty, and, I suspect there might be parts requiring more modification. Nevertheless, it, almost gives the entire construction. $b)$ Let $\{a_n\}_{n=1}^\infty$ be defined as $a_n = a+nb$, for every $n\in\mathbb{N}$, where $a,b$ fixed. Suppose that, $a=da_1, b=db_1, (a_1,b_1)=1$. $a_n = d(a_1 +nb_1)$. Since, $(a_1,b_1)=1$, Dirichlet's theorem on arithmetic progressions tell us that, the sequence $\tilde{a}_n = a_1 + nb_1$ has infinitely many prime numbers. Just, select $n$ to be large enough, so that $a_1+nb_1>d$. Hence, there exists (in fact, infinitely many) primes $p$'s such that, $a_n = pd$ for some $d<p$, thus, it cannot be a prime power.

After about two years from my initial post, here is a shorter solution to part $({\rm b})$ of this problem. Take any arbitrary positive arithmetic progression $a_1,\dots,a_N$ with common difference $d>0$. The key observation is that if $(a_1,\dots,a_N)$ is an arithmetic progression, so do $(Ta_1,\dots,Ta_N)$. Equipped with this, we now prescribe $T$ as follows. We search for a $T$ of form $T=a_1^{\alpha_1}\cdots a_N^{\alpha_N}$, where the exponents are to be determined shortly. With this, now let $b_i=Ta_i$. Then, $b_i = a_i\prod_{j=1}^{N}a_j^{\alpha_j}$. Now comes the prescription of the exponents. Let $q_1<q_2<\cdots<q_N$ be fixed prime numbers. We select the sequence as follows: for every $1\leqslant i\leqslant N$, $\alpha_i\equiv -1\pmod{q_i}$. For every $j\neq i$, $\alpha_j\equiv 0\pmod{q_i}$. By the Chinese remainder theorem, such numbers do exist. Finally, this yields $b_i$ is a perfect $q_i^{th}$ power for every $i$, concluding the construction.