On the question about $\sin 30^\circ$ and $\tan 30^\circ$, I don't know why the post says that you do not understand how to do this. You have in fact given a full justification of the fact that $\sin 30^\circ <\tan 30^\circ$.
You wrote, correctly, that $\sin 30^\circ=\frac{y}{r}$ and $\tan 30^\circ=\frac{y}{x}$, and that $r>x$. So when you divide $y$ by $r$, you must get something smaller than when you divide $y$ by $x$. The desired result follows.
That is probably the best way to view things. But you may also know that $\tan \theta=\frac{\sin\theta}{\cos \theta}$. So to get $\tan\theta$, you divide $\sin\theta$ by $\cos\theta$. Let's assume that $\cos\theta$ is positive. (This is true as long as $\theta<90^\circ$.) Note that $\cos\theta <1$, because $x$ is always less than the hypotenuse $r$. When you divide a number $a$ by a positive number $b<1$, the answer is $>a$.
For the second problem, the idea was fine, but there was a problem with the details. We will show that $\cos 27^\circ$ is less than $\cos 26^\circ$. How one explains it depends on how you visualize the trigonometric functions. If you increase the angle from $26^\circ$ to $27^\circ$, keeping $x$ unchanged, then (draw a picture!) "$r$" will increase, so the cosine $\frac{x}{r}$ will decrease.
Or else keep $r$ constant at $1$. Then as your angle increases from $26^\circ$ to $27^\circ$, the number "$x$" will decrease, so cosine will decrease. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\cos\theta$ steadily decreases. It starts at $1$ and ends up at $0$.
A roughly similar argument shows that as $\theta$ increases from $0^\circ$ to $90^\circ$, $\sin\theta$ steadily increases. It starts at $0$ and ends up at $1$.
The behaviour of $\tan\theta$ is much wilder. As $\theta$ increases from $0^\circ$ to $90^\circ$, $\tan\theta$ steadily increases, after a while very rapidly. It becomes enormously large as $\theta$ approaches $90^\circ$.
For the long run, it will be very useful to have a good mental image of the shape of the curves $y=\sin x$, $y=\cos x$, and $y=\tan x$.