1. No, if you want to have a 1D defect in $d$ spatial dimensions, you need $\pi_{d-2} (M)$ to be nontrivial.

2. The fact that your Lagrangian has nontrivial $\pi_{1} (M)$ implies, when extended to 4 spatial dimensions, that there are 2D defects.

1. No, if you want to have a 1D defect in $d$ spatial dimensions, you need $\pi_{d-2} (M)$ to be nontrivial.

2. The fact that your Lagrangian has nontrivial $\pi_{1} (M)$ implies, when extended to 4 spatial dimensions, that there are 2D defects.

AMENDED to address OP's follow-up question:

3. If you want a 1D defect in 4 spatial dimensions, you need to use a model that yields a nontrivial $\pi_{2} (M)$ (which would yield a 0D defect in 3 spatial dimensions, such as a magnetic monopole). Typically you'd use something with an internal $SU(2)$ symmetry, see for example this discussion.