= Solution
The <Seifert-van Kampen theorem> states that if a space is covered by path-connected open sets $U,V$ with path-connected intersection containing a basepoint, then its <fundamental group> is the <free product> $\pi_1(U)*\pi_1(V)$ modulo the <normal subgroup> generated by identifying the two images of each loop in $U\cap V$. More generally it is the corresponding pushout of groups.
A wedge of two circles has <fundamental group> $F(a,b)$, the <free group> on their two loops: apply the theorem to thickened circle neighborhoods whose intersection is <contractible> and use $\pi_1(S^1)=\mathbb Z$. Represent the <torus> by a square with opposite edges identified. Remove an interior disk; the remainder deformation retracts onto this wedge. Replacing the disk attaches a two-cell whose boundary follows $aba^{-1}b^{-1}$. A disk neighborhood and the punctured <torus> have annular intersection, and van Kampen kills precisely that boundary loop. Hence
$$
\boxed{\pi_1(T)=\langle a,b\mid aba^{-1}b^{-1}=1\rangle\cong\mathbb Z^2.}
$$
A <torus> with a point removed retracts to a <torus> with a small open disk removed, and then to the same wedge. Thus
$$
\boxed{\pi_1(T\setminus\{p\})\cong F_2.}
$$
For the <connected sum>, remove a disk from each <torus> and glue their boundary circles with reversed orientations. The two pieces have free groups $F(a,b)$ and $F(c,d)$ and their overlap is an annulus. Its boundary generators correspond to $[a,b]$ and $[c,d]^{-1}$, where $[a,b]=aba^{-1}b^{-1}$. Therefore
$$
\boxed{\pi_1(T\#T)=\langle a,b,c,d\mid[a,b][c,d]=1\rangle.}
$$
The boundary relation is essential: this group is neither the <free group> on four generators nor the abelian group $\mathbb Z^4$.
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