Problem 178

(1) Since vectors $\overrightarrow {a}$ and $\overrightarrow {b}$ satisfy $| \overrightarrow {a}|=| \overrightarrow {b}|=1$, and $|2 \overrightarrow {a}- \overrightarrow {b}|= \sqrt {5}$,
$5=|2 \overrightarrow {a}- \overrightarrow {b}|^{2}=4 \overrightarrow {a}^{2}+ \overrightarrow {b}^{2}-4 \overrightarrow {a}\cdot \overrightarrow {b}=4×1^{2}+1^{2}-4 \overrightarrow {a}\cdot \overrightarrow {b}$,
$\overrightarrow {a}\cdot \overrightarrow {b}=0$.
$| \overrightarrow {2a}- \overrightarrow {3b}|= \sqrt {4 \overrightarrow {a}^{2}+9 \overrightarrow {b}^{2}-12 \overrightarrow {a}\cdot \overrightarrow {b}}= \sqrt {4×1^{2}+9×1^{2}-0}= \sqrt {13}$.

(2)
$|3 \overrightarrow {a}- \overrightarrow {b}|= \sqrt {9 \overrightarrow {a}^{2}+ \overrightarrow {b}^{2}-6 \overrightarrow {a}\cdot \overrightarrow {b}}= \sqrt {9+1-0}= \sqrt {10}$,
$| \overrightarrow {a}-2 \overrightarrow {b}|= \sqrt { \overrightarrow {a}^{2}+4 \overrightarrow {b}^{2}-4 \overrightarrow {a}\cdot \overrightarrow {b}}= \sqrt {1+4-0}= \sqrt {5}$,
$(3 \overrightarrow {a}- \overrightarrow {b})\cdot ( \overrightarrow {a}-2 \overrightarrow {b})=3 \overrightarrow {a}^{2}+2 \overrightarrow {b}^{2}-7 \overrightarrow {a}\cdot \overrightarrow {b}=5$.
$\cos θ= \frac {(3 \overrightarrow {a}- \overrightarrow {b})\cdot ( \overrightarrow {a}-2 \overrightarrow {b})}{|3 \overrightarrow {a}- \overrightarrow {b}| | \overrightarrow {a}-2 \overrightarrow {b}|}= \frac {5}{ \sqrt {10}× \sqrt {5}}= \frac { \sqrt {2}}{2}$,
Since θ∈[0,π], $θ= \boxed{\frac {π}{4}}$.