Answer
$a.$ The slope is $m_{tan} = \frac{1}{4}$.
$b.$ The equation is $y=\frac{1}{4}x+\frac{7}{4}.$
Work Step by Step
$a.$ Using the formula for the slope of the tangent at point $P(a,f(a))$ from definition (2) with $a=1$ and $f(a)=2$ we have
$$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\sqrt{(1+h)+3}-2}{h}=\lim_{h\to0}\frac{\sqrt{h+4}-2}{h}\cdot\frac{\sqrt{h+4}+2}{\sqrt{h+4}+2}=\lim_{h\to0}\frac{\sqrt{h+4}^2-2^2}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{h+4-4}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{h}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{1}{\sqrt{h+4}+2}=\frac{1}{\sqrt{0+4}+2}=\frac{1}{4}.$$
$b.$ Using the formula from definition (1) for the equation of the slope $y-f(a)=m_{tan}(x-a)$ with the same values for $a$ and $f(a)$ as in the part $a.$ and the calculated value $m_tan=\frac{1}{4}$ we get
$$y-2=\frac{1}{4}(x-1)\Rightarrow y-2=\frac{1}{4}x-\frac{1}{4},$$
which gives
$$y=\frac{1}{4}x-\frac{1}{4}+2=\frac{1}{4}x+\frac{7}{4}.$$
