1. **State the problem:** We need to find the possible values of $a$ such that the constant term in the expansion of $$\left(\frac{x^2}{2} + \frac{a}{x}\right)^6$$ is 960. 2. **Formula and approach:** The binomial expansion of $\left(A + B\right)^n$ is given by $$\sum_{k=0}^n \binom{n}{k} A^{n-k} B^k$$ where $A = \frac{x^2}{2}$ and $B = \frac{a}{x}$. 3. **Find the general term:** The $k$-th term is $$T_{k+1} = \binom{6}{k} \left(\frac{x^2}{2}\right)^{6-k} \left(\frac{a}{x}\right)^k = \binom{6}{k} \frac{x^{2(6-k)}}{2^{6-k}} \cdot a^k x^{-k} = \binom{6}{k} \frac{a^k}{2^{6-k}} x^{12 - 2k - k} = \binom{6}{k} \frac{a^k}{2^{6-k}} x^{12 - 3k}$$ 4. **Find the constant term:** The constant term occurs when the power of $x$ is zero: $$12 - 3k = 0 \implies 3k = 12 \implies k = 4$$ 5. **Calculate the constant term:** Substitute $k=4$: $$T_5 = \binom{6}{4} \frac{a^4}{2^{6-4}} x^0 = \binom{6}{4} \frac{a^4}{2^2} = 15 \cdot \frac{a^4}{4} = \frac{15}{4} a^4$$ 6. **Set equal to 960 and solve for $a$:** $$\frac{15}{4} a^4 = 960 \implies a^4 = \frac{960 \times 4}{15} = \frac{3840}{15} = 256$$ 7. **Find $a$:** $$a = \pm \sqrt[4]{256} = \pm 4$$ **Final answer:** The possible values of $a$ are $\boxed{4 \text{ and } -4}$.