Ребята пожалуйста ПОМОГИТЕ очень СРОЧНО надо номер 3.5

Решите задачу:

$\tt 1) \ \ \sqrt{122^2-22^2}= \sqrt{(122+22)(122-22)}=\sqrt{144\cdot100}=\\\\ {}\ \ \ =\sqrt{144}\cdot \sqrt{100} =12\cdot10=120\\\\\\ 2) \ \ \sqrt{6.8^2-3.2^2}= \sqrt{(6.8+3.2)(6.8-3.2)}=\sqrt{10\cdot3.6}=\\\\ {}\ \ \ =\sqrt{36}=6$

$\displaystyle\tt 3) \ \ \sqrt{45.8^2-44.2^2}= \sqrt{(45.8+44.2)(45.8-44.2)}=\sqrt{90\cdot1.6}=\\\\ {}\ \ \ =\sqrt{144}=12\\\\\\ 4) \ \ \sqrt{3.13^2-3.12^2}= \sqrt{(3.13+3.12)(3.13-3.12)}=\sqrt{6.25\cdot 0.01}=\\\\ {}\ \ \ =\sqrt{\frac{625}{100} }\cdot\sqrt{\frac{1}{100} }=\frac{25}{10}\cdot \frac{1}{10}= \frac{1}{4}= 0.25$