Answer
$a)$ see the proof.
$b)$ the concentration is near to $30$ so for large time, the concentration of the solution in the tank will be the same as the concentration of the brine.
Work Step by Step
$a$-
Since the variation of the mass in the reaction is zero it follows that:
$$dm_s=0 \to m_b=m_{salt in the tank}$$
Before the tank we have:
$$dm_b=C_b \cdot dV_b$$
$$dm_b=C_b \cdot Q_bdt$$
$$m_b=C_b \cdot Q_bt$$
In the tank:
$$m_{salt in the tank}=C_s(t)\cdot V_T $$
$$m_{salt in the tank}=C_s(t)\cdot (V_w+V_b(t))$$
$$m_{salt in the tank}=C_s(t)\cdot (V_w+Q_bt)$$
so:
$$C_b \cdot Q_bt=C_s(t)\cdot (V_w+Q_bt)$$
$$30 \cdot 25t=C_s(t)\cdot (5000+25t)$$
$$\frac{30 \cdot 25t}{5000+25t}=C_s(t)$$
$$\frac{30t}{200+t}=C_s(t)$$
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$b$-
$$\lim\limits_{t \to \infty}\frac{30t}{200+t}=\lim\limits_{t \to \infty}\frac{30t}{t(\frac{200}{t}+1)}=\lim\limits_{t \to \infty}\frac{30}{\frac{200}{t}+1}=30$$
The concentration is near to $30$ so for large time, the concentration of the solution in the tank will be the same as the concentration of the brine.