Is non-adaptive CPA strictly weaker than standard CPA security?
Edit. Except for CPA1 (defined by the OP), "CPA" below refers to multi-challenge CPA, where the adversary always submits two messages and receive the encryption of either the first or the second messages. It's equivalent to the usual one-challenge multi-query CPA.

CPA1 is equivalent to CPA. Consider real-or-random CPA (RRCPA), where the adversary adaptively chooses messages and receive either encryptions of the chosen messages or encryptions of random messages. It is routine to see that CPA is tightly equivalent to RRCPA. Lastly, RRCPA reduces to CPA1 via a standard hybrid argument, as the encryption of random messages can be queried before the challenge. The proof is not tight, though.

In more detail, let $\mathcal{A}$ be an adversary against RRCPA, we construct $\mathcal{B}$ against CPA1. Let $Q$ be a polynomial that upper-bounds the number of challenges $\mathcal{A}$ makes. $\mathcal{B}$ does the following:


Sample $i\gets[Q]$ and launch $\mathcal{A}$.
For the first ${(i-1)}$ challenges $m_1,\dots,m_{i-1}$ that $\mathcal{A}$ makes, query the encryption oracle to obtain the correct ciphertexts $c_1,\dots,c_{i-1}$ and use them as the response to $\mathcal{A}$.
Sample ${Q-(i-1)}$ random messages $m_i',\dots,m_Q'$ and query the encryption oracle to get the ciphertexts $c_{i+1},\dots,c_Q$ (without $c_i$).
When $\mathcal{A}$ makes the $i$th challenge $m_i$, submit $m_i,m_i'$ to the CPA1 challenger and gets back a ciphertext $c_i$. Use it to respond to $\mathcal{A}$.
For the other challenges from $\mathcal{A}$, use $c_{i+1},\dots,c_Q$ as the responses.


Let $H_i$ be the hybrid of RRCPA such that the first $i$ challenges are real and the other are random, and $G_b$ the CPA1 experiments.
Clearly, $\mathcal{A}$ is in $H_{i-1+b}$ if $\mathcal{B}$ is in $G_b$ and chooses $i$ at the beginning.
Averaging over $i$ shows that the advantage of $\mathcal{B}$ is $1/Q$ that of $\mathcal{A}$.
This finishes the reduction.