<s>[INST] Find the CDF if $P(X=1)=\frac{1}{2}$ and $P(a
The Problem: Let $X$ be a random variable that satisfies $P(X=1)=\frac{1}{2}$ and $P(a<X<b|X\ne1)=\frac{b-a}{2}$ for $0\leq a<b\leq2$. Find the cumulative distribution function of $X$.
My Thoughts: We have that
$$P(0<X<2)=P(0<X<2|X\ne1)P(X\ne1)+P(0<X<2|X=1)P(X=1)=1.$$
Therefore, $P(X\leq t)=0$ for $t\leq0$, and $P(X\leq t)=1$ for $t\geq2$. Now if $0<t<1$ then
\begin{equation}\begin{split}
P(0<X<2)
&=P(X=t)+P(0<X<t)+P(t<X<1)+P(X=1)+P(1<X<2)\\
&=P(X=t)+\frac{t}{4}+\frac{1-t}{4}+\frac{1}{2}+\frac{1}{4}\\
&=P(X=t)+1\\
&=P(X=t)+P(0<X<2),
\end{split}\end{equation}
so that $P(X=t)=0$. Similarly, $P(X=t)=0$ for $1<t<2$.
Next, we note that for $0<t<1$ we have
$$P(X\leq t)=P(0<X<t|X=1)P(X=1)+P(0<X<t|X\ne1)P(X\ne1)=\frac{t}{4}.$$
If $t=1$, then
$$P(X\leq1)=\frac{1}{2}+P(0<X<1)=\frac{1}{2}+P(0<X<1|X\ne1)P(X\ne1)=\frac{3}{4}.$$
Finally, if $1<t<2$ then
$$P(X\leq t)=P(0<X\leq1)+P(1<X\leq t)=\frac{3}{4}+P(1<X<t|X\ne1)=\frac{3}{4}+\frac{t-1}{8}.$$
Therefore, we have
$$F_X(t)=\begin{cases}
0&\text{if }t\leq0\\
\dfrac{t}{4}&\text{if }0<t<1\\
\dfrac{3}{4}+\dfrac{t-1}{4}&\text{if }1\leq t<2\\
1&\text{if }t\geq2.
\end{cases}$$

 [/INST] Your end result is okay.
Let me add a route that makes things less error prone. 

Let $Y$ have uniform distribution on interval $[0,2]$ and let $Z$ be a random variable degenerated at $1$ in the sence that $P(Z=1)=1$.
Then we find:
$$F_{X}\left(x\right)=P\left(X=1\right)P\left(X\leq x\mid X=1\right)+P\left(X\neq1\right)P\left(X\leq x\mid X\neq1\right)=\frac{1}{2}F_{Z}\left(x\right)+\frac{1}{2}F_{Y}\left(x\right)$$
It can easily be checked that the RHS  is the same as your final result.
Here $F_Z(x)=\mathbf{1}_{\left[1,\infty\right)}\left(x\right)$ and $F_{Y}\left(x\right)=\begin{cases}
0 & \text{if }x\leq0\\
\frac{1}{2}x & \text{if }0<x<2\\
1 & \text{otherwise}
\end{cases}$
</s>