= Solution
For one visitor, let
$$
T_0=0,
\qquad
T_j=Z_1+\cdots+Z_j,
$$
and define
$$
p_j(u)=\mathbb P(T_{j-1}\leq u<T_j),
\qquad 1\leq j\leq10.
$$
A visitor arriving at time $s\leq t$ is in room $j$ at time $t$ exactly when its independent service-time mark satisfies $T_{j-1}\leq t-s<T_j$. Thus it receives mark $j$ with probability $p_j(t-s)$; a further mark records that it has already left the gallery.
The <Independent marking theorem for Poisson point processes> makes the point processes carrying the different room marks independent. Consequently $V_1(t),\ldots,V_{10}(t)$ are independent, and <Poisson thinning theorem>[Poisson thinning] gives
$$
V_j(t)\sim\operatorname{Poisson}(m_j(t)),
\qquad
m_j(t)=\int_0^t\lambda(s)p_j(t-s)\,ds.
$$
Equivalently,
$$
m_j(t)=\int_0^t\lambda(s)
\mathbb P\left(
Z_1+\cdots+Z_{j-1}\leq t-s
<Z_1+\cdots+Z_j
\right)ds.
$$
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