2 . Given the parametric equation of line $1_{1}$ as
$$
\left\{\begin{array}{l}
x=\frac{2}{3} \sqrt{3} t-1, \\
y=\frac{1}{3} \sqrt{3} t+1,
\end{array}\right.
$$

and the polar equation of line $1_{2}$ as
$$
\rho \sin \left(\theta-\frac{\pi}{4}\right)=2 \text {. }
$$

Find the degree measure of the acute angle $\alpha$ between line $I_{1}$ and $I_{2}$.
$2.1_{1}$ can be transformed into the equation $\mathrm{y}=\frac{1}{2} \mathrm{x}+\frac{3}{2}$, so $\mathrm{k}_{1}=\frac{1}{2}$, and the equation of 2 can be transformed into $\mathrm{y}=\mathrm{x}+2 \sqrt{2}$, so $\mathrm{k}_{2}=1$.
$$
\begin{array}{l}
\text { Thus } \quad \operatorname{tg} \alpha=\left|\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}\right|=\left|\frac{1-\frac{1}{2}}{1+1 \times \frac{1}{2}}\right|=\frac{1}{3}, \\
\therefore \quad \alpha=\operatorname{arctg} \frac{1}{3} .
\end{array}
$$