7. Given real numbers $a, b$ satisfy $\left(a-\frac{b}{2}\right)^{2}=1-\frac{7}{4} b^{2}$, let $t_{\max }$ and $t_{\min }$ be the maximum and minimum values of $t=a^{2}+2 b^{2}$, respectively, then $t_{\text {max }}+t_{\text {min }}=$ $\qquad$ .
7. From $\left(a-\frac{b}{2}\right)^{2}=1-\frac{7}{4} b^{2}$, we know $a^{2}+2 b^{2}=a b+1$.

From $\frac{a^{2}+2 b^{2}}{2 \sqrt{2}}=\frac{|a|^{2}+2\left|b^{2}\right|}{2 \sqrt{2}} \geqslant \frac{2 \sqrt{2}|a b|}{2 \sqrt{2}}=|a b|$, we know $-\frac{a^{2}+2 b^{2}}{2 \sqrt{2}} \leqslant a b \leqslant \frac{a^{2}+2 b^{2}}{2 \sqrt{2}},-\frac{a^{2}+2 b^{2}}{2 \sqrt{2}}+1 \leqslant a b+1 \leqslant$ $\frac{a^{2}+2 b^{2}}{2 \sqrt{2}}+1$,

i.e., $-\frac{t}{2 \sqrt{2}}+1 \leqslant t \leqslant-\frac{t}{2 \sqrt{2}}+1$.
From this, we solve $\frac{2 \sqrt{2}}{2 \sqrt{2}+1} \leqslant t \leqslant \frac{2 \sqrt{2}}{2 \sqrt{2}-1}$, i.e., $\frac{8-2 \sqrt{2}}{7} \leqslant t \leqslant \frac{8+2 \sqrt{2}}{7}$, with equality holding if and only if $a=\sqrt{2} b$ and $a=-\sqrt{2} b$ respectively.

Thus, $t_{\max }+t_{\min }=\frac{8+2 \sqrt{2}}{7}+\frac{8-2 \sqrt{2}}{7}=\frac{16}{7}$.