1) The set ${t + 1, t2 + 2, t2 + t}$ is a basis of $F_3[t]≤2$. I put false because if t is 2, then we have ${t + 1, 0 , t2 + t}$ so a non zero coefficient could exist.
2) T : V → V a linear transformation of a vector space V. If $ker T ∩ Im T = {0}$ then T is injective. Im not sure about this at all.
3) Let $A ∈ M_{p×n}$ and $B ∈ M_{p×1}$. $AX = B$ a system of unknowns $X1, . . . , Xn$. If the rank of matrix A is p, then the system has at least 1 solution.
On
To a)
It follows from the answer by @MattS that it is false.
To b)
Consider $T=0.$ Then, $\mathrm{im}(T)=\{0\},$ from where $\mathrm{ker}(T)\cap\mathrm{im}(T)=\{0\}.$ Now, $ \mathrm{ker}(T)=V.$ Thus, if $V\ne \{0\}$ $T$ is not injective.
To c)
It is $\mathrm{rank}(A)=\mathrm{rank}(A|B)=p.$ So using the theorem of Rouche-Capelli we get that there is at least one solution.
The answer to 1) is false. To prove this, note that $$t+1-(t^2+t)+(t^2+2)=0$$ assuming $F_3$ is the field with 3 elements.
2) is also false. Consider the map $\mathbb{R}^2\to \mathbb{R}^2$ sending $(x,y)\mapsto (x,0)$.
3) is true. We may use Gaussian elimination to prove this.