fix: 消息日志增强 + 历史消息抑制 + SSE实时追踪 + 群聊上下文优化

- 日志：收/发消息均显示群名称，管理员显示真实QQ昵称而非"开拓者"
- 历史消息：服务重启后NapCat回放的历史消息不再触发回复，静默注入上下文
- 消息时间戳：转发给AI时附带【消息时间: HH:MM:SS (XmXs前)】标记
- ♪ 分割符：QQ消息支持♪作为句子断点
- AI-Core SSE端点：全链路追踪实时推送，ethend不再5秒轮询
- 群聊上下文：AI-Core明确被告知消息来自群聊，以实际发送者为主语

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

backend/ai-core/internal/llm/call_log.go

@@ -52,6 +52,8 @@ func (cl *CallLogger) log(r CallRecord) {
 	if cl.size < cl.capacity {
 		cl.size++
 	}
+
+	broadcastCall(r)
 }
 
 func (cl *CallLogger) get(limit int) []CallRecord {
@@ -72,3 +74,49 @@ func (cl *CallLogger) get(limit int) []CallRecord {
 	}
 	return result
 }
+
+// --- SSE subscriber system ---
+
+type callSubscriber struct {
+	ch   chan CallRecord
+	done chan struct{}
+}
+
+var (
+	callSubscribers   []*callSubscriber
+	callSubscribersMu sync.RWMutex
+)
+
+// SubscribeCalls returns a channel that receives new CallRecords and a done channel.
+func SubscribeCalls() (<-chan CallRecord, <-chan struct{}) {
+	ch := make(chan CallRecord, 20)
+	done := make(chan struct{})
+	callSubscribersMu.Lock()
+	callSubscribers = append(callSubscribers, &callSubscriber{ch: ch, done: done})
+	callSubscribersMu.Unlock()
+	return ch, done
+}
+
+// UnsubscribeCalls removes a subscriber. Safe to call multiple times.
+func UnsubscribeCalls(ch <-chan CallRecord) {
+	callSubscribersMu.Lock()
+	defer callSubscribersMu.Unlock()
+	for i, s := range callSubscribers {
+		if s.ch == ch {
+			close(s.done)
+			callSubscribers = append(callSubscribers[:i], callSubscribers[i+1:]...)
+			return
+		}
+	}
+}
+
+func broadcastCall(r CallRecord) {
+	callSubscribersMu.RLock()
+	defer callSubscribersMu.RUnlock()
+	for _, s := range callSubscribers {
+		select {
+		case s.ch <- r:
+		default:
+		}
+	}
+}