Question: I need help with equation $y''(x)-y'(x)tgx+2y(x)=0$ and $y(0)=1,y'(0)=1$

We can guess solution $y\_1(x)=\sin x$ and assume that second solution is $y\_2(x)=z(x)\sin x$. To find $z(x)$ we have to put our assumed solution into equation.
Putting into equation and multiplying by $\cos x$ we get:
$$
\sin x\cos x\cdot z''(x)+(2\cos^2x-\sin^2x)z'(x)=0
$$
We can substitute $z'=g$ and obtain after little work:
$$\frac{dg}{g}=\frac{\sin^2x-\cos^2x}{\sin x\cos x}dx$$
Now there is assumption in work that $\sin x\neq0$, so $x\neq0$. Therefore we won't find anything useful for solution which is defined in $0$. So do we need to carry on with that way, or we can just safely assume that $\sin x=0$ and have $$2z'(x)=0$$ therefore $z=const$ so solution which work is $y=\sin x +1$?
Do I need to worry about uniqueness of this solution?

Answer: 
The case $z'=0$ leads back to the family $y=C\sin x$ of solutions that you already know. $y=1$ and thus also $y=\sin x+1$ are not solutions of the equation.

You lost a factor of $2$ while transforming.

The integration of the correct identity should lead to
$$
z'=\frac{C}{\sin^2x\cos x}=\frac{C\cos x}{\sin^2x(1-\sin^2x)}.
$$

As you have a linear ODE, uniqueness of the solutions follows from the continuity of the coefficients. $\tan x$ is continuous on a neighborhood of $x=0$.
The solution you have obtained satisfies the initial conditions so it is indeed unique.