（1）由题意可知乙队以$3:0$或$3:1$取胜，

当乙队以$3:0$获胜时，${{P}_{1}}={{\left(\dfrac{1}{3}\right)}^{3}}=\dfrac{1}{27}$，

当乙队以$3:1$获胜时，${{P}_{2}}={\rm C}_{3}^{1}\times \dfrac{2}{3}\times {{\left(\dfrac{1}{3}\right)}^{2}}\times \left(\dfrac{1}{3}\right)=\dfrac{2}{27}$，

$\therefore $ 甲、乙两队比赛$1$场后，乙队积（$3$分）的概率为$P={{P}_{1}}+{{P}_{2}}=\dfrac{1}{27}+\dfrac{2}{27}=\dfrac{1}{9}$；

（2）记“甲、乙比赛两场后，两队积分相等”为事件$A$，

设第$i$场甲、乙两队积分分别为${{X}_{i}}$，${{Y}_{i}}$，

则${{X}_{i}}=3-{{Y}_{i}}$，$i=1$，$2$，

因两队积分相等，

$\therefore {{X}_{1}}+{{X}_{2}}={{Y}_{1}}+{{Y}_{2}}$，

即${{X}_{1}}+{{X}_{2}}=(3-{{X}_{1}})+(3-{{X}_{2}})$，则${{X}_{1}}+{{X}_{2}}=3$，

而$P(X=0)={{\left(\dfrac{1}{3}\right)}^{3}}+{\rm C}_{3}^{1}\times \dfrac{2}{3}\times {{\left(\dfrac{1}{3}\right)}^{2}}\times \dfrac{1}{3}=\dfrac{1}{9}$，

$P(X=1)={\rm C}_{4}^{2}\times {{\left(\dfrac{2}{3}\right)}^{2}}\times {{\left(\dfrac{1}{3}\right)}^{2}}\times \dfrac{1}{3}=\dfrac{8}{81}$，

$P(X=2)={\rm C}_{4}^{2}\times {{\left(\dfrac{2}{3}\right)}^{2}}\times {{\left(\dfrac{1}{3}\right)}^{2}}\times \dfrac{2}{3}=\dfrac{16}{81}$，

$P(X=3)={\rm C}_{3}^{2}\times {{\left(\dfrac{2}{3}\right)}^{2}}\times \dfrac{1}{3}\times \dfrac{2}{3}+{{\left(\dfrac{2}{3}\right)}^{3}}=\dfrac{16}{27}$

$\therefore P(A)=P({{X}_{1}}=0)P({{X}_{2}}=3)+P({{X}_{1}}=1)P({{X}_{2}}=2)+P({{X}_{1}}=2)P({{X}_{2}}=1)+P({{X}_{1}}=3)P({{X}_{2}}=0)$

$=\dfrac{1}{9}\times \dfrac{16}{27}+\dfrac{8}{81}\times \dfrac{16}{81}+\dfrac{16}{81}\times \dfrac{8}{81}+\dfrac{16}{27}\times \dfrac{1}{9}$

$=\dfrac{1120}{6561}$．