Writing vectors in the $\hat{r}$ and $\hat{\theta}$ basis I am having a bit of confusion about writing down vectors in the $\hat{r}$,  $\hat{\theta}$ basis. Let's say I want to write $-1$ $\hat{i}$ $+$ $1$ $\hat{j}$ in the $\hat{r}$,  $\hat{\theta}$ basis. Could I write $1$ $\hat{r}$ $+$ $1$ $\hat{\theta}$ so long as I specify that $\hat{r}$ is "at" $\pi/2$ radians? (...and in the above question $\hat{\theta}$ would be "at" $\pi$ radians)

Could I have described the exact same point with $\sqrt2\hat{r}$ $+$ $0$ $\hat{\theta}$ so long as I specify this time that $\hat{r}$ is "at" $3\pi/4$ radians? (...and $\hat{\theta}$ would be "at" $5\pi/4$ radians, but we wouldn't move along the $\hat{\theta}$  direction this time).
Yes, you can do so.

Remember the vectors $\hat r,\hat\theta$ are defined by the equations
$$\begin{cases}\hat r=\cos\theta\hat i+\sin\theta\hat j\\ \hat\theta=-\sin\theta\hat i+\cos\theta\hat j.\end{cases}\tag{1}$$

This is equivalent to $$\begin{cases}\hat i=\cos\theta\hat r-\sin\theta\hat\theta\\ \hat j=\sin\theta\hat r+\cos\theta\hat\theta.\end{cases}\tag{2}$$

To see why this is true, remember that each of $\hat r,\hat\theta$ are obtained by rotating each of $\hat i,\hat j$ by $\theta$ radian anti-clockwise (which is what $(1)$ does). So we can rotate each of $\hat r,\hat\theta$ by $-\theta$ radian anti-clockwise to obtain $\hat i,\hat j$ (which is what $(2)$ does).

So $-1\hat i+1\hat j=-(\cos\theta\hat r-\sin\theta\hat\theta)+(\sin\theta\hat r+\cos\theta\hat\theta)=(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta$. Plug in the value $\theta$ and you can express $-1\hat i+1\hat j$ in terms of $\hat r,\hat\theta$.

For case 1, you put $\theta=\frac{\pi}{2}$, giving $(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta=\hat r+\hat\theta$. For case 2, you put $\theta=\frac{3\pi}{4}$, giving $(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta=\sqrt2\hat r$.
You need a combination. You are correct in saying that your value of $\hat{r}$ needs to be $\sqrt{2}$, but by saying $0\hat{\theta}$ you are saying that your point lies in the $\hat{i}$ direction. Saying that it is "at" $3\pi /4$ radians is precisely what $3\pi /4 \hat{\theta}$ means.

Your answer in polar coordinates $(r, \theta)$ should be $\left(\sqrt2 , \frac{3\pi}{4}\right).$

Think of it in terms of $re^{i\theta} = r\cos\theta\hat{i} + r\sin\theta\hat{j}$