1) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,x_3,\ldots)$ (which is your idea for (4)). Then $T$ is one to one, because if $T(x)=T(y)$, then $(0,x_1,x_2,x_3,\ldots)=(0,y_1,y_2,y_3\ldots)$ and hence $x_i=y_i$ for each $i\in\mathbb N$. But $T$ is not onto, since there is no sequence $x\in l^p$ such that $T(x)=(1,0,0,0,0,\ldots)$.
2) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(x_2,x_3,\ldots)$. Then $T$ is onto, since if $y=(y_1,y_2,y_3,\ldots)\in l^p$ is given, then choose $x=(0,y_1,y_2,y_3,\ldots)$ to obtain $T(x)=y$. But $T$ is not one to one, since $T((0,0,0,0,\ldots))=T((1,0,0,0,0,0,\ldots))$.
3) There is no such $T$, since if $T$ is bijective, then it is invertible, and since $T$ is linear, its inverse must be linear as well. For the linearity of the inverse, see, for instance, Is the inverse of a linear transformation linear as well?
4) Take your favorite infinite-dimensional vector space $V$ and consider $T(x):=0$ for each $x\in V$. Then $T$ is neither one to one nor onto, which should be obvious.