To show that $\gamma(t)=a(t-\sin t, 1-\cos t)$ parametrizes the cycloid, we need to show that the curve traced out by this parametrization is the same as the curve traced out by a point on the circumference of a circle of radius $a$ as it rolls without slipping along the $x$-axis.

Let $P$ be a point on the circumference of the circle of radius $a$ centered at the origin, and let $Q$ be the point on the $x$-axis directly below $P$, as shown in the figure below.

Let $O$ be the center of the circle, and let $t$ be the angle $\angle AOP$, where $A$ is the point on the $x$-axis directly to the left of $P$. Then the angle $\angle AOQ$ is also $t$, since $OA$ and $OQ$ are radii of the circle.

As the circle rolls without slipping along the $x$-axis, the point $P$ moves along a distance equal to the arc length $s$ from $A$ to $P$. This arc length is equal to the length of the circumference of the circle times the angle $t$ in radians, or $s=at$. Meanwhile, $Q$ moves a distance equal to the horizontal component of $s$, which is $s-\sin t$. Thus, the position of $Q$ at time $t$ is $(s-\sin t, 0)=(at-a\sin t, 0)$. Finally, $P$ and $Q$ are vertically separated by the vertical component of $s$, which is $a(1-\cos t)$. Thus, the position of $P$ at time $t$ is $(at-a\sin t, a(1-\cos t))$.

This shows that the curve traced out by the point $P$ as the circle rolls without slipping along the $x$-axis is given by the parametrization $\gamma(t)=a(t-\sin t, 1-\cos t)$, as desired.

(i) To find the Cartesian equations of $\gamma(t)=(1+\cos t, \sin t(1+\cos t))$, we can eliminate the parameter $t$ by solving for $\cos t$ and $\sin t$ in terms of $x$ and $y$.

From the first coordinate, we have $x=1+\cos t$, so $\cos t=x-1$.

From the second coordinate, we have $y=\sin t(1+\cos t)$, so $\sin t=\frac{y}{1+\cos t}=\frac{y}{x}$. Substituting for $\cos t$ and $\sin t$ in terms of $x$ and $y$, we get

$\begin{aligned} & x=1+\cos t=1+(x-1)=x, \ & y=\sin t(1+\cos t)=\frac{y}{x} \cdot x(x-1)=y(x-1) .\end{aligned}$

Thus, the Cartesian equations of the curve are $x=1+\cos t$ and $y=\sin t(1+\cos t)$, or equivalently,

$x=1+\frac{x^2-y^2}{2 x}, \quad y=\frac{y}{x} \cdot\left(1+\frac{x^2-y^2}{2 x}\right)$.

Simplifying, we get

$x^2-2 x+y^2=1$,

which is the equation of a circle centered at $(1,0)$ with radius $1$.

(ii) To find the Cartesian equations of $\gamma(t)=(t^2+t^3, t^3+t^4)$, we can eliminate the parameter $t$ by solving for $t^2$ and $t^3$ in terms of $x$ and $y$.

From the first coordinate, we have $x=t^2+t^3=t^2(1+t)$, so $t^2=\frac{x}{1+t}$.

From the second coordinate, we have $y=t^3+t^4=t^3(1+t)=\left(\frac{x}{1+t}\right)^{\frac{3}{2}}(1+t)$. Substituting for $t^2$ and $t^3$ in terms of $x$ and $y$, we get

$\begin{aligned} t^2 & =\frac{x}{1+t} \ t^3 & =\frac{y}{(1+t)^{\frac{3}{2}}}\end{aligned}$

Solving for $t$ in terms of $x$ and $y$, we find

$t=\frac{\sqrt{x^2+4 y}-x}{2}$

Substituting this expression for $t$ into the equation for $x$, we get

$\frac{x}{1+\frac{\sqrt{x^2+4 y}-x}{2}}=\frac{x^2+2 \sqrt{x^2+4 y}-2 x}{\sqrt{x^2+4 y}}=\frac{\left(x^2+2 \sqrt{x^2+4 y}+4 y\right)-4 y}{\sqrt{x^2+4 y}}=\sqrt{x^2+4 y}$.

To find the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis, we need to find the $t$-values where $\gamma(t)$ intersects the $x$-axis, and then integrate the magnitude of the derivative of $\gamma(t)$ over the interval of $t$ values corresponding to that part of the curve.

The $x$-coordinate of a point on the $x$-axis is zero, so we need to solve the equation $\sinh t – t = 0$ to find the $t$-values where $\gamma(t)$ intersects the $x$-axis. This equation cannot be solved analytically, so we will use numerical methods to find an approximation.

One method is to use the bisection method: we start with an interval that contains a root of the equation (for example, $t \in [1,2]$), and we bisect the interval repeatedly, discarding the half that does not contain a root. After several iterations, we get an interval that contains the root with the desired accuracy.

Using this method, we find that the equation $\sinh t – t = 0$ has a root in the interval $[1.5,1.6]$, which we will denote by $t_0$. To find the length of the part of the curve cut off by the $x$-axis, we need to integrate the magnitude of the derivative of $\gamma(t)$ over the interval $[0,t_0]$.

The derivative of $\gamma(t)$ is

$\gamma^{\prime}(t)=(\cosh t-1, \sinh t)$

and its magnitude is

$\left|\gamma^{\prime}(t)\right|=\sqrt{(\cosh t-1)^2+(\sinh t)^2}=\sqrt{\cosh ^2 t-2 \cosh t+2}$

To integrate this expression over the interval $[0,t_0]$, we make the substitution $u = \sinh t$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} \sqrt{\cosh ^2\left(\sinh ^{-1} u\right)-2 \cosh \left(\sinh ^{-1} u\right)+2} d u$.

Using the identity $\cosh^2 x – \sinh^2 x = 1$, we can simplify the expression under the square root to $2 – 2 \sinh(\sinh^{-1} u) = 2(1 – u^2)$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} 2 \sqrt{1-u^2} d u=\pi$

Therefore, the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis is $\boxed{\pi}$.