Vectorfield $F(x,y)=-yi+4xj$, expres this vectorfield in vectors r and $\theta$. With $\hat{r}=\cos(\theta)i+\sin(\theta)j$ and $\hat{\theta}=-\sin(\theta)i+\cos(\theta)j$.
So I wrote $F(x,y)$ in polar coordinates, which gives $r(-\sin(\theta)i+4\cos(\theta)j)$, I don't know how to write this as $\hat{r}$ and $\hat{\theta}$ because of that 4 in front of the cosine.
Like you wrote:
$$ \hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j} $$ $$ \hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j} $$
One can conclude (by projecting, or by just solving these equations) that:
$$ \hat{i} = \cos \theta \hat{r} - \sin \theta \hat{\theta} $$ $$ \hat{j} = \sin \theta \hat{r} + \cos \theta \hat{\theta} $$
Meaning that
$$F(r,\theta)=-r\sin \theta \hat{i} + 4r\cos \theta \hat{j} = $$ $$(-r\sin \theta \cos \theta + 4r \cos \theta \sin \theta)\hat{r} + (r\sin^2{\theta} + 4r \cos^2 {\theta})\hat{\theta} = $$ $$ \frac{3}{2} r \sin (2\theta) \hat{r} + r(3\cos^2 \theta +1)\hat{\theta}$$