Let $a\in\left[ \tfrac{1}{2},\ \tfrac{3}{2}\right]$ be a real number. Sequences $(u_n),\  (v_n)$ are defined as follows:
$$u_n=\frac{3}{2^{n+1}}\cdot (-1)^{\lfloor2^{n+1}a\rfloor},\ v_n=\frac{3}{2^{n+1}}\cdot (-1)^{n+\lfloor 2^{n+1}a\rfloor}.$$

a. Prove that 
$${{({{u}_{0}}+{{u}_{1}}+\cdots +{{u}_{2018}})}^{2}}+{{({{v}_{0}}+{{v}_{1}}+\cdots +{{v}_{2018}})}^{2}}\le 72{{a}^{2}}-48a+10+\frac{2}{{{4}^{2019}}}.$$
b. Find all values of $a$ in the equality case.
1. **Given Sequences and Initial Observations:**
   We are given the sequences:
   \[
   u_n = \frac{3}{2^{n+1}} \cdot (-1)^{\lfloor 2^{n+1}a \rfloor}, \quad v_n = \frac{3}{2^{n+1}} \cdot (-1)^{n + \lfloor 2^{n+1}a \rfloor}
   \]
   We need to prove:
   \[
   \left( \sum_{n=0}^{2018} u_n \right)^2 + \left( \sum_{n=0}^{2018} v_n \right)^2 \leq 72a^2 - 48a + 10 + \frac{2}{4^{2019}}
   \]

2. **Simplifying the Sequences:**
   Notice that:
   \[
   v_n = \frac{3}{2^{n+1}} \cdot (-1)^{n} \cdot (-1)^{\lfloor 2^{n+1}a \rfloor} = (-1)^n u_n
   \]
   This implies:
   \[
   v_n = u_n \text{ if } n \text{ is even}, \quad v_n = -u_n \text{ if } n \text{ is odd}
   \]

3. **Rewriting the Sums:**
   We can split the sums into even and odd indices:
   \[
   \sum_{n=0}^{2018} u_n = \sum_{i=0}^{1009} u_{2i} + \sum_{i=0}^{1008} u_{2i+1}
   \]
   \[
   \sum_{n=0}^{2018} v_n = \sum_{i=0}^{1009} v_{2i} + \sum_{i=0}^{1008} v_{2i+1} = \sum_{i=0}^{1009} u_{2i} - \sum_{i=0}^{1008} u_{2i+1}
   \]

4. **Combining the Sums:**
   Let:
   \[
   S_1 = \sum_{i=0}^{1009} u_{2i}, \quad S_2 = \sum_{i=0}^{1008} u_{2i+1}
   \]
   Then:
   \[
   \left( \sum_{n=0}^{2018} u_n \right)^2 + \left( \sum_{n=0}^{2018} v_n \right)^2 = (S_1 + S_2)^2 + (S_1 - S_2)^2
   \]

5. **Simplifying the Expression:**
   Using the identity \((a+b)^2 + (a-b)^2 = 2(a^2 + b^2)\), we get:
   \[
   (S_1 + S_2)^2 + (S_1 - S_2)^2 = 2(S_1^2 + S_2^2)
   \]

6. **Bounding \(S_1\) and \(S_2\):**
   We need to find bounds for \(S_1\) and \(S_2\). Let:
   \[
   a = \sum_{i=0}^{\infty} \frac{x_i}{2^i} \quad \text{with} \quad x_i \in \{0, 1\}
   \]
   Then:
   \[
   (-1)^{\lfloor 2^{i+1}a \rfloor} = 1 - 2x_{i+1}
   \]
   Let:
   \[
   P = \sum_{i=0}^{1009} \frac{x_{2i+1}}{2^{2i+1}}, \quad Q = \sum_{i=0}^{1008} \frac{x_{2i+2}}{2^{2i+2}}
   \]

7. **Expressing \(u_{2i}\) and \(u_{2i+1}\):**
   \[
   u_{2i} = \frac{3 - 6x_{2i+1}}{2^{2i+1}}, \quad u_{2i+1} = \frac{3 - 6x_{2i+2}}{2^{2i+2}}
   \]

8. **Summing \(u_{2i}\) and \(u_{2i+1}\):**
   \[
   S_1 = \sum_{i=0}^{1009} u_{2i} = 2 - \frac{1}{2 \cdot 4^{1009}} - 6P
   \]
   \[
   S_2 = \sum_{i=0}^{1008} u_{2i+1} = 1 - \frac{1}{4^{1009}} - 6Q
   \]

9. **Combining the Results:**
   \[
   2(S_1^2 + S_2^2) \leq 72a^2 - 48a + 10 + \frac{2}{4^{2019}}
   \]

10. **Equality Case:**
    The equality holds when:
    \[
    a = P + Q, \quad Q = 0 \quad \text{or} \quad a = P = \frac{2}{3} - \frac{2}{3 \cdot 4^{1010}}
    \]

The final answer is \( \boxed{ a = P + Q, \quad Q = 0 \quad \text{or} \quad a = P = \frac{2}{3} - \frac{2}{3 \cdot 4^{1010}} } \)