50б даю за решение, срочнякккк

Решите задачу:

$1)\; \; \sqrt{3-2\sqrt{2}}\cdot (1+\sqrt2)= \sqrt{1+2-2\cdot 1\cdot \sqrt2} \cdot (1+\sqrt2)=\\\\= \sqrt{1^2+(\sqrt2)^2-2\cdot 1\cdot \sqrt2} \cdot (1+\sqrt2)=\sqrt{(1-\sqrt2)^2}\cdot (1+\sqrt2)=\\\\=|1-\sqrt2|\cdot (1+\sqrt2)=(\sqrt2-1)\cdot (\sqrt2+1)=(\sqrt2)^2-1^2=2-1=1\\\\\\\star \; \; 1-\sqrt2\approx1-1,4=-0,4\ \textless \ 0\; \; \Rightarrow \; \; |1-\sqrt2|=-(1-\sqrt2)=\\\\=\sqrt2-1\; \; \star$

$2)\; \; b=-1 \frac{1}{8} \\\\(b+4)^2-(b-3)(b+3)=b^2+8b+16-(b^2-9)=\\\\=8b+25=8\cdot (-1\frac{1}{8})+25=-8\cdot \frac{9}{8}+25=-9+25=16$