Why does the host ask for the poppy address by arp if it already knows it?

Question

Computer networks

In wireshark, I was able to capture such an arp request:

Here you can see that f6 with address 15 asks about who is at address 3, but does this not by broadcasting, but right at f1 itself (address 3). That is, as if he wants to make sure that address 3 is still on poppy f1.
Who knows why? OS f6-win10 f1-debian9

Answer 1

Because there is DHCP and all that (by "all that" you should understand APIPA\Link-local)
https://datatracker.ietf.org/doc/html/rfc1122

The ARP specification [LINK:2] suggests but does not
require a timeout mechanism to invalidate cache entries
when hosts change their Ethernet addresses. The
prevalence of proxy ARP (see Section 2.4 of [INTRO:2])
has significantly increased the likelihood that cache
entries in hosts will become invalid, and therefore
some ARP-cache invalidation mechanism is now required
for hosts. Even in the absence of proxy ARP, a long-
period cache timeout is useful in order to
automatically correct any bad ARP data that might have
been cached.

:

Approaches (1) and (2) involve ARP cache timeouts on
the order of a minute or less.

https://ru.wikipedia.org/wiki/ARP

The effectiveness of ARP functioning depends largely on the ARP cache (ARP cache) that is available on each host. The cache contains a table of correspondence between MAC and IP addresses compiled by the operating system.
The lifetime of a cache entry is left to the discretion of the developer. The default can be from tens of seconds (for example, 20 seconds) to four hours (Cisco IOS).

Answer 2

The request is of course sent by a broadcaster. Expand the ARP header - you will see zeros there in the Target MAC address field .
The packet where Who has is written is the request, the one in which is at is the answer, and the answer, of course, is sent to the requester, and not to everyone in a row.