22. Solution: (1) From the problem conditions, $|a| = 1$. Thus the parametric equation of $l$ is $\left\{\begin{array}{l} x = 4t + 1 \\ y = 3t - 1 \end{array}\right.$ ($t$ is the parameter). The parametric equation of circle $C$ is $\left\{\begin{array}{l} x = 1 + \cos\theta \\ y = -2 + \sin\theta \end{array}\right.$ ($\theta$ is the parameter). & \hline \end{tabular} Eliminating parameter $t$, the ordinary equation of $l$ is $3x - 4y - 7 = 0$. ..... 3 marks Eliminating parameter $\theta$, the ordinary equation of $C$ is $(x-1)^2 + (y+2)^2 = 1$. ..... 5 marks (2) The equation of $l'$ is $y = \frac{3}{4}(x + m) - \frac{7}{4}$, i.e., $3x - 4y + 3m - 7 = 0$. ..... 6 marks Since circle $C$ has only one point at distance $\mathbf{1}$ from $l'$, and the radius of circle $C$ is $\mathbf{1}$, the distance from $C(1, -2)$ to $l'$ is 2. ..... 8 marks That is, $\frac{|3 + 8 + 3m - 7|}{5} = 2$. Solving, we get $m = 2$ ($m = -\frac{14}{3} < 0$ is rejected). ..... 10 marks
22. Solution: (1) From the problem conditions, $|a| = 1$. Thus the parametric equation of $l$ is $\left\{\begin{array}{l} x = 4t + 1 \\ y = 3t - 1 \end{array}\right.$ ($t$ is the parameter). The parametric equation of circle $C$ is $\left\{\begin{array}{l} x = 1 + \cos\theta \\ y = -2 + \sin\theta \end{array}\right.$ ($\theta$ is the parameter). & \\
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Eliminating parameter $t$, the ordinary equation of $l$ is $3x - 4y - 7 = 0$. ..... 3 marks
Eliminating parameter $\theta$, the ordinary equation of $C$ is $(x-1)^2 + (y+2)^2 = 1$. ..... 5 marks
(2) The equation of $l'$ is $y = \frac{3}{4}(x + m) - \frac{7}{4}$, i.e., $3x - 4y + 3m - 7 = 0$. ..... 6 marks
Since circle $C$ has only one point at distance $\mathbf{1}$ from $l'$, and the radius of circle $C$ is $\mathbf{1}$, the distance from $C(1, -2)$ to $l'$ is 2. ..... 8 marks
That is, $\frac{|3 + 8 + 3m - 7|}{5} = 2$. Solving, we get $m = 2$ ($m = -\frac{14}{3} < 0$ is rejected). ..... 10 marks