Let's say we have a queue with size n, for example 30, no item will be added to the queue, the size of the queue is fixed, arrival rate is zero. There is one server with shifted exponential service distribution, for example it takes at least 1 minute for the server to give the service to the each item in the queue. In addition, on average 2 added minutes the server needs for each item. It means the $PDF$ of service distribution is: $\frac{1}{2} e^{-1/2(t-1)}$
1 - What is probability the at certain time $t_1$, exactly $k$ items out of n items have been serviced?
2 - By time time $t_1$, we expect how many items have been serviced?
Denote by $S_k=X_1+\ldots+X_k=k+Y_1+\ldots+Y_k$ the total service time of first $k$ customers. Here $Y_i$ have PDF $\frac12 e^{-\frac12t}\mathbb 1_{\{t>0\}}$. Let $Z_k=Y_1+\ldots+Y_k$.
The first question asks on the probability that $S_k\leq t_1$, $S_{k+1}>t_1$. Write it: $$ \mathbb P(S_k\leq t_1, \ S_{k+1}>t_1)=\mathbb P(S_k\leq t_1) -\mathbb P( S_{k+1}\leq t_1) $$ $$=\mathbb P(Z_k\leq t_1-k) - \mathbb P(Z_{k+1}\leq t_1-k-1) $$ Next use that $Z_m$ is Gamma-distributed (or Erlang distributed) with PDF $$f_{Z_m}(s)=\frac{1}{2^m(m-1)!} s^{m-1}e^{-\frac12s}\mathbb 1_{\{s>0\}}.$$
The desired probability equals to the difference of CDF's of $Z_k$ and $Z_{k+1}$ in the points $t_1-k$ and $t_1-k-1$ respectively: $$ F_{Z_k}(t_1-k)-F_{Z_{k+1}}(t_1-k-1) $$ It can be expressed as a difference of two sums, see https://en.wikipedia.org/wiki/Erlang_distribution#Cumulative_distribution_function_(CDF).
To find expectation of number $N$ of served customers up to time $t_1$, one do not need to find $\mathbb P(N=k)$ which is calculated above. Let us use the first step only: $$ \mathbb P(N=k)=\mathbb P(S_k\leq t_1, \ S_{k+1}>t_1)=\mathbb P(S_k\leq t_1) -\mathbb P( S_{k+1}\leq t_1) $$ Next, $$ \mathbb E[N] = \sum_{k=1}^\infty k\mathbb P(N=k) $$ $$ = 1\cdot(\mathbb P(S_1\leq t_1) -\mathbb P( S_2\leq t_1)) $$ $$+2\cdot(\mathbb P(S_2\leq t_1) -\mathbb P( S_3\leq t_1)) $$ $$+3\cdot(\mathbb P(S_3\leq t_1) -\mathbb P( S_4\leq t_1)) + \ldots $$ $$ =\sum_{k=1}^{\infty}\mathbb P(S_k\leq t_1)=\sum_{k=1}^{\infty}\mathbb P(Z_k\leq t_1-k)=\sum_{k=1}^{\lfloor{t_1}\rfloor}\int\limits_0^{t_1-k} \frac{1}{2^k(k-1)!} t^{k-1}e^{-\frac12t}\,dt. $$ Doubt that it is possible to simplify this integral.