$(1)$ $\overset{\large\rightharpoonup}{u} = \overset{\large\rightharpoonup}{v} + (2,2) \implies | \overset{\large\rightharpoonup}{u}+\overset{\large\rightharpoonup}{v} | = | 2\overset{\large\rightharpoonup}{v} + (2,2) | = 2 | \overset{\large\rightharpoonup}{v} + (1,1) | = 2 \implies | \overset{\large\rightharpoonup}{v} + (1,1) | = 1$，對。 $(2)$ $| \overset{\large\rightharpoonup}{u}+\overset{\large\rightharpoonup}{v} |^2 - | \overset{\large\rightharpoonup}{u}-\overset{\large\rightharpoonup}{v} |^2 = 4\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v} \implies 4 - 8 = 4\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v} \implies \overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v} = -1$，對。 $(3)$ $\overset{\large\rightharpoonup}{v}$ 終點在以 $(-1,-1)$ 為圓心、$1$ 為半徑的圓上，與原點最遠距離為 $\sqrt{2} + 1 > 2$，故 $|\overset{\large\rightharpoonup}{v}|$ 可以大於 $2$，錯。 $(4)$ 若反向平行且 $\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v} = -1 \implies |\overset{\large\rightharpoonup}{u}||\overset{\large\rightharpoonup}{v}| = 1$，又 $|\overset{\large\rightharpoonup}{u}|^2 + |\overset{\large\rightharpoonup}{v}|^2 = 6$，此時長度有實數解，故可能平行，錯。 $(5)$ 二維外積大小為 $|u_x v_y - u_y v_x| = 2|v_y - v_x|$，因 $\overset{\large\rightharpoonup}{v}$ 滿足 $(x+1)^2 + (y+1)^2 = 1$，最高差 $|y-x|$ 最大值為 $\sqrt{2}$，故 $|\overset{\large\rightharpoonup}{u}||\overset{\large\rightharpoonup}{v}|\sin \theta \le 2\sqrt{2} \implies \sin \theta \le \dfrac{2\sqrt{2}}{|\overset{\large\rightharpoonup}{u}||\overset{\large\rightharpoonup}{v}|}$，對。 故選 $(1)(2)(5)$。